Chapter 16, Problem 16.127QA

Interpretation Introduction

To find:

In the titration of pyridine and $HCl$, answer the following questions.

a) The pH at equivalence point.

b) The best choice indicator for this titration.

c) The pH for the titration after different volumes.

d) Sketch the titration curve for this titration and indicate which major species are present in the solution when pH = pKa at equivalence point and at the end of titration.

Answer to Problem 16.127QA

Solution:

a) pH at equivalence point = $2.95$

b) Methyl orange

c) At $0.0$ mL pH $= 9.40$, at $1.0$ mL pH =$6.26$, at $3.0$ mL pH =$5.69$, at $7.0$ mL pH =$5.05$, at $9.0$ mL pH =$4.69$, at $10.0$ mL pH =$4.43$, at $10.5$ mL pH =$4.25$, at $11.0$ mL pH =$3.96$, at $11.5$ mL pH =$3.16$, at $11.8$ mL pH =$2.43$, at $12.5$ mL pH =$1.79$, at $15.0$ mL pH =$1.26$, at $20.0$ mL pH =$0.93$.

d)

Explanation of Solution

1) Concept:

To calculate the pH of a weak base with a strong acid, we need to use reaction stoichiometry and RICE table. First, we will calculate the equivalence point from the volume and molarity of base and acid. We will calculate the pH at equivalence point by using RICE table. At equivalence point, moles of pyridine and HCl are the same, so the only species present is protonated pyridine. Using the hydrolysis of protonated pyridine and ICE table we can find the pH at the equivalence point. Suitable indicator for this titration is decided by comparing the pH at equivalence point and the pH range of indicators given in Figure 16.5. We will sketch the titration curve based on the pH of the solution and mL of  a strong acid added to the weak base.

2) Formula:

i) $pH = - \log \left[H_3{O}^{+}\right]$ 

ii) $p{K}_a=-\log \left[K_a\right]$ 

iii) $K_a=\frac{\left[A^{-}\right]\left[H_3{O}^{+}\right]}{\left[HA\right]}$

iv) $K_a={10}^{-p{K}_a}$

3) Given:

i) Concentration of pyridine =$367 mM = 367 mM \times \frac{1 M}{1000 mM}=0.367 M$

ii) Volume of pyridine = $15.8 mL=0.0158 L$

iii) Concentration of HCl = $0.500 M$

iv) $p{K}_b=8.77$

v) $K_b={10}^{-8.77}=1.70 \times {10}^{-9}$

vi) $K_a=\frac{K_w}{K_b}= \frac{1.0 \times {10}^{-14}}{1.70 \times {10}^{-9}}=5.88 \times {10}^{-6}$

4) Calculations:

a) To calculate the equivalence point of the titration between pyridine and $HCl$, first we will write the reaction and calculate the volume of $HCl$.

$C_5{H}_{5}N \left(aq\right)+HCl \left(aq\right)\to C_5{H}_5{NH}^{+}\left(aq\right)+C{l}^{-} \left(aq\right)$

From the balanced chemical equation, the mole ratio between pyridine and HCl is $1:1$.

At the equivalence point, moles of pyridine = moles of $HCl$

$n_{pyridine}= 0.367 M \times 0.0158 L= 0.0057986 mol pyridine$

Moles of $HCl$ at the equivalence point $=0.0057986 mol$

Volume of $HCl$ required to reach the equivalence point

$V_{HCl}= \frac{0.0057986}{0.500 M}= 0.0115972 L=11.60 mL$

The volume of $HCl$ required to reach the equivalence point $=11.60 mL$

At equivalence point the number of moles of a weak base is equal to the number of moles of a strong acid. So, at the end of the reaction, the solution contains protonated pyridine which will govern the $pH$ of the solution at the equivalence point.

	$C_5{H}_{5}N \left(aq\right)+HCl \left(aq\right)\to C_5{H}_5{NH}^{+}\left(aq\right)+C{l}^{-} \left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$	$HCl \left(aq\right)$	$C_5{H}_5{NH}^{+}\left(aq\right)$
Initial	$0.0057986 mol$	$0.0057986 mol$	$0$
Added	$- 0.0057986 mol$	$- 0.0057986 mol$	$+ 0.0057986 mol$
Final	$0$	$0$	$0.0057986 mol$

We have the moles of the conjugate acid, hence, we can calculate the pH of the solution using the RICE table for protonated pyridine.

The dissociation reaction of protonated pyridine (conjugate acid) is written as:

$C_5{H}_5{NH}^{+}\left(aq\right)+H_{2}O \left(l\right) \rightleftharpoons C_5{H}_{5}N \left(aq\right)+H_3{O}^{+}\left(aq\right)$

Now calculating new molarity of conjugate acid as:

$M=\frac{0.0057986 mol}{\left(0.0158+0.01160\right)L }=0.212 M$

We can use RICE table to calculate the pH of the solution of conjugate acid.

	$C_5{H}_5{NH}^{+}\left(aq\right)+H2O \left(l\right) \rightleftharpoons C_5{H}_{5}N \left(aq\right)+H_3{O}^{+}\left(aq\right)$
	$C_5{H}_5{NH}^{+}\left(aq\right)$(M)	$C_5{H}_{5}N \left(aq\right)$(M)	$H_3{O}^{+}\left(aq\right)$(M)
Initial	$0.212$	$0$	$0$
Change	$-x$	$+x$	$+x$
Change	$0.212-x$	$x$	$x$

Now write the $K_a$ expression to calculate the $\left[H_3{O}^{+}\right].$

$K_a=\frac{\left[C_5{H}_{5}N\right]\left[H_3{O}^{+}\right]}{\left[C_5{H}_5{NH}^{+}\right]}$

$5.88 \times {10}^{-6} =\frac{\left(x\right)\left(x\right)}{(0.212-x)}$

We will apply the 5% rule here, so we can ignore x from it, $0.212-x=0.212$

$5.88 \times {10}^{-6} =\frac{\left(x\right)\left(x\right)}{\left(0.212\right)}$

$x^2=1.2466 \times {10}^{-6}$

$x=0.001116$

$x= \left[H_3{O}^{+}\right]= 0.001116$

$pH= -\log \left[H_3{O}^{+}\right]= -\log \left(0.001116\right)=2.95$

The pH of the solution at equivalence point = $2.95$

b) The pH of the solution at equivalence point =$2.95$, we can use methyl orange as an indicator for this titration.

c) Calculate the pH for the titration after following the volume of HCl added to a weak base.

1) $0.0$mL HCl added to a weak base.

We can use RICE table to calculate the pH of the solution of pyridine at 0.0 mL addition of HCl

	$C_5{H}_{5}N \left(aq\right)+H_{2}O \left(l\right) \rightleftharpoons C_5{H}_5{NH}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
	$C_5{H}_{5}N\left(aq\right)$(M)	$C_5{H}_5{NH}^{+} \left(aq\right)$(M)	$O{H}^{-}\left(aq\right)$(M)
Initial	$0.367$	$0$	$0$
Change	$-x$	$+x$	$+x$
Equilibrium	$0.367-x$	$x$	$x$

Now write the $K_b$ expression to calculate the $\left[O{H}^{-}\right].$

$K_b=\frac{\left[C_5{H}_5{NH}^{+}\right]\left[O{H}^{-}\right]}{\left[C_5{H}_{5}N\right]}$

$1.70 \times {10}^{-9} =\frac{\left(x\right)\left(x\right)}{(0.367-x)}$

We will apply the 5% rule here, so we can ignore x from it,$0.367-x=0.367$,

$1.70 \times {10}^{-9} =\frac{\left(x\right)\left(x\right)}{\left(0.367\right)}$

$x^2=6.27 \times {10}^{-10}$

$x=2.498 \times {10}^{-5}$

$x= \left[O{H}^{-}\right]= 2.498 \times {10}^{-5}$

$pOH= -\log \left[O{H}^{-}\right]= -\log \left(2.498 \times {10}^{-5}\right)=4.60$

$pH =14- pOH$

$pH =14- 4.60$

$pH =9.40$

Initial pH of the pyridine  = $9.40$

2) $1.0$mL HCl added to a weak base.

Calculate the moles of pyridine from 15.80 mL and 0.367 M.

$0.0158 L \times \frac{0.367 mol}{1 L} = 0.0057986 mol$

Calculate the mole of HCl, when 1.0 mL added to the weak base.

$1.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.000500 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.000500$	$0$
Change	$-0.000500$	$-0.000500$	$+0.000500$
Final	$0.005299$	$0$	$0.000500$

The total sample volume is $15.8 mL + 1.0 mL = 16.8 mL or 0.0168 L$ and the concentrations of the base and conjugate acid is

$\left[C_5{H}_{5}N\right]= \frac{0.005299 mol}{0.0168 L}=0.3154 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.000500 mol}{0.0168 L}=0.02976 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.029762\right)}{\left(0.3154\right)} =8.77+\log \left(0.097365\right)=8.77-1.025191=7.7448$

$pH=14-7.7448=6.26$

3) $3.0$mL HCl added to a weak base.

Calculate the mole of HCl, when $3.0$ mL is added to the weak base.

$3.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0015 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0015$	$0$
Change	$-0.0015$	$-0.0015$	$+0.0015$
Final	$0.004299$	$0$	$0.0015$

The total sample volume is $15.8 mL + 3.0 mL = 18.8 mL or 0.0188 L$ and the concentrations of the base and conjugate acid are

$\left[C_5{H}_{5}N\right]= \frac{0.004299 mol}{0.0188 L}=0.2286 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.0015 mol}{0.0188 L}=0.07979 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.07979\right)}{\left(0.2286\right)} =8.77+\log \left(0.3490\right)=8.77-0.45717=8.3128$

$pH=14-8.3128=5.69$

4) $7.0$mL HCl added to a weak base.

Calculate the mole of HCl, when $7.0$ mL is added to the weak base.

$7.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0035 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$  are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0035$	$0$
Change	$-0.0035$	$-0.0035$	$+0.0035$
Final	$0.002299$	$0$	$0.0035$

The total sample volume is $15.8 mL + 7.0 mL = 22.8 mL or 0.0228 L$ and the concentrations of the base and conjugate acid are

$\left[C_5{H}_{5}N\right]= \frac{0.002299 mol}{0.0228 L}=0.1008 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.0035 mol}{0.0228 L}=0.1535 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.1535\right)}{\left(0.1008\right)} =8.77+\log \left(1.5228\right)=8.77+0.183=8.95$

$pH=14-8.95=5.05$

5) $9.0$mL HCl added to a weak base.

Calculate the mole of HCl, when $9.0$ mL is added to the weak base.

$9.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0045 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0045$	$0$
Change	$-0.0045$	$-0.0045$	$+0.0045$
Final	$0.001299$	$0$	$0.0045$

The total sample volume is $15.8 mL + 9.0 mL = 24.8 mL or 0.0248 L$ and the concentrations of the base and conjugate acid are

$\left[C_5{H}_{5}N\right]= \frac{0.001299 mol}{0.0248 L}=0.0524 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.0045 mol}{0.0248 L}=0.1815 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.1815\right)}{\left(0.0524\right)} =8.77+\log \left(3.46374\right)=8.77+0.54=9.31$

$pH=14-9.31=4.69$

6) $10.0$mL HCl added to a weak base.

Calculate the mole of HCl, when $10.0$ mL isadded to the weak base.

$10.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0050 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0050$	$0$
Change	$-0.0050$	$-0.0050$	$+0.0050$
Final	$0.000799$	$0$	$0.0050$

The total sample volume is $15.8 mL + 10.0 mL = 25.8 mL or 0.0258 L$ and the concentrations of the base and conjugate acid are

$\left[C_5{H}_{5}N\right]= \frac{0.000799 mol}{0.0258 L}=0.0310 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.0050 mol}{0.0258 L}=0.1938 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.1938\right)}{\left(0.0310\right)} =8.77+\log \left(6.2516\right)=8.77+0.796=9.57$

$pH=14-9.57=4.43$

7) $10.5$mL HCl added to a weak base.

Calculate the mole of HCl, when $10.5$ mL is added to the weak base.

$10.5 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0053 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0053$	$0$
Change	$-0.0053$	$-0.0053$	$+0.0053$
Final	$0.000549$	$0$	$0.0053$

The total sample volume is $15.8 mL + 10.5 mL = 26.3 mL or 0.0263L$ and the concentrations of the base and conjugate acid is

$\left[C_5{H}_{5}N\right]= \frac{0.000549 mol}{0.0263 L}=0.0209 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.0053 mol}{0.0263 L}=0.1996 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.1996\right)}{\left(0.0209\right)} =8.77+\log \left(9.55024\right)=8.77+0.98=9.75$

$pH=14-9.75=4.25$

8) $11.0$mL HCl added to a weak base.

Calculate the mole of HCl, when $11.0$ mL ia added to the weak base.

$11.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0055 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right)\to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0055$	$0$
Change	$-0.0055$	$-0.0055$	$+0.0055$
Final	$0.000299$	$0$	$0.0055$

The total sample volume is $15.8 mL + 11.0 mL = 26.8 mL or 0.0268 L$ and the concentrations of the base and conjugate acid are

$\left[C_5{H}_{5}N\right]= \frac{0.000299 mol}{0.0268 L}=0.0111 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.0055 mol}{0.0268 L}=0.2052 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.2052\right)}{\left(0.0111\right)} =8.77+\log \left(18.4193\right)=8.77+1.267=10.04$

$pH=14-10.04=3.96$

9) $11.5$mL HCl added to a weak base.

Calculate the mole of HCl, when $11.5$ mL is added to the weak base.

$11.5 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.00575 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.00575$	$0$
Change	$-0.00575$	$-0.00575$	$+0.00575$
Final	$0.0000486$	$0$	$0.00575$

The total sample volume is $15.8 mL + 11.5 mL = 27.3 mL or 0.0273 L$ and the concentrations of the base and conjugate acid are

$\left[C_5{H}_{5}N\right]= \frac{0.0000486 mol}{0.0273 L}=0.0018 M$

$\left[C_5{H}_5{NH}^{+}\right]= \frac{0.00575 mol}{0.0273L}=0.2106 M$

Using these values in the Henderson–Hasselbalch equation gives us

$pOH=p{K}_b+\log \frac{\left[C_5{H}_5{NH}^{+}\right]}{\left[C_5{H}_{5}N\right]}$

$pOH=8.77+\log \frac{\left(0.2106\right)}{\left(0.0018\right)} =8.77+\log \left(117\right)=8.77+2.068=10.84$

$pH=14-10.84=3.16$

10) $11.8$mL HCl added to a weak base.

Calculate the mole of HCl, when $11.8$ mL is added to the weak base.

$11.8 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0059 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0059$	$0$
Change	$-0.0057986$	$-0.0057986$	$+0.0057986$
Final	$0$	$0.0001014$	$0.0057986$

HCl is a strong acid, the pH of solution can be calculated from the concentration of HCl.

The total sample volume is $15.8 mL + 11.8 mL = 27.6 mL or 0.0276 L$.

$\left[HCl\right]= \frac{0.0001014 mol}{0.0276 L}=0.0036739 M$

$pH= -\log \left[H_3{O}^{+}\right]= -\log \left(0.0036739\right)=2.43$

11) $12.5$mL HCl added to a weak base.

Calculate the mole of HCl, when $12.5$ mL is added to the weak base.

$12.5 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.00625 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.00625$	$0$
Change	$-0.0057986$	$-0.0057986$	$+0.0057986$
Final	$0$	$0.0004514$	$0.0057986$

HCl is a strong acid, the pH of solution can be calculated from concentration of HCl.

The total sample volume is $15.8 mL + 12.5 mL = 28.3 mL or 0.0283 L$.

$\left[HCl\right]= \frac{0.0004514mol}{0.0283 L}=0.00160 M$

$pH= -\log \left[H_3{O}^{+}\right]= -\log \left(0.00160\right)=1.79$

12) $15.0$mL HCl added to a weak base.

Calculate the mole of HCl, when $15.0$ mL is added to the weak base.

$15.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.0075 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.0075$	$0$
Change	$-0.0057986$	$-0.0057986$	$+0.0057986$
Final	$0$	$0.0017014$	$0.0057986$

HCl is a strong acid, the pH of solution can be calculated from concentration of HCl.

The total sample volume is $15.8 mL + 15.0 mL = 30.8 mL or 0.0308 L$.

$\left[HCl\right]= \frac{0.0017014 mol}{0.0308 L}=0.0552 M$

$pH= -\log \left[H_3{O}^{+}\right]= -\log \left(0.0552\right)=1.26$

13) $20.0$mL HCl added to a weak base.

Calculate the mole of HCl, when $20.0$ mL is added to the weak base.

$20.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.500 mol}{1 L} = 0.010 mol HCl$

We can use a modified RICE table to determine the how many moles of $C_5{H}_{5}N$ remain and how many moles of $C_5{H}_5{NH}^{+}$ are produced.

	$C_5{H}_{5}N\left(aq\right)+HCl \left(l\right) \to C_5{H}_5{NH}^{+} \left(aq\right)+C{l}^{-}\left(aq\right)$
	$C_5{H}_{5}N \left(aq\right)$(mol)	$HCl \left(aq\right)$(mol)	$C_5{H}_5{NH}^{+}\left(aq\right)$(mol)
Initial	$0.0057986$	$0.010$	$0$
Change	$-0.0057986$	$-0.0057986$	$+0.0057986$
Final	$0$	$0.0042014$	$0.0057986$

HCl is a strong acid, the pH of solution can be calculated from concentration of HCl.

The total sample volume is $15.8 mL + 20.0 mL = 35.8 mL or 0.0358 L$.

$\left[HCl\right]= \frac{0.0042014 mol}{0.0358 L}=0.1174 M$

$pH= -\log \left[H_3{O}^{+}\right]= -\log \left(0.1174\right)=0.93$

d) Sketch the titration curve (pH vs mL of strong acid added) for this titration and show the major species are present in solution at pH = pka, equivalence point and end point.

Conclusion:

The pH of the solution is calculated from reaction stoichiometry and RICE table at different mL of the titrant added to the weak base and also calculated at equivalence point. Sketch drawn on the basis of pH values and volume of strong acid added to a weak base.