   # To have a buffer with a pH of 2.50, what volume of 0.150 M NaOH must be added to 100. mL of 0.230 M H 3 PO 4 ? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 109IL
Textbook Problem
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## To have a buffer with a pH of 2.50, what volume of 0.150 M NaOH must be added to 100. mL of 0.230 M H3PO4?

Interpretation Introduction

Interpretation:

The volume of 0.150 M, NaOH is to be calculated when it is mixed with 100 mL 0.230 M,H3PO4 to get the buffer solution of pH value equals to 2.50.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of H3PO4 with NaOH. The equilibrium can be represented as,

H3PO4(aq)+NaOH(aq)H2O(l)+NaH2PO4(aq)

Calculation of pH at various points is done as follows,

The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution H3PO4/NaH2PO4. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, when concentration of acid and its conjugate base are equal pH value at midpoint will be given as;

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

### Explanation of Solution

The volume of NaOH used is calculated below.

Given:

Refer to section 17.3  in the textbook for the value of Ka.

The value of Ka for H3PO4 is 7.5×103.

The pKa value is calculated as follows;

pKa=log(Ka)

Substitute, 7.5×103 for Ka.

pKa=log(7.5×103)=2.12

Therefore, pKa value is 2.12.

The initial concentration of H3PO4 is 0.230 molL1.

The initial concentration of NaOH is 0.150 molL1.

The volume of H3PO4 is 100mL.

Conversion of 100 mL into L.

(100mL)(1L1000mL)=0.100L

Let the volume of NaOH added xmL.

Conversion of xmL into L.

(xmL)(1L1000mL)=0.00xL

The total volume after the reaction is calculated as,

totalvolume=volumeof NaHC2O4(L) + volume of NaOH(L)

totalvolume = 0.100(L)+0.00x(L)=(0.100+0.00x)L

Therefore, total volume after reaction is (0.100+0.00x)L.

The calculation of moles is done by using the expression,

Numberof moles=concentration(molL1)volume(L)

The ICE table (1) for the reaction between NaOH and H3PO4 is given below,

EquationH3PO4(aq)+NaOH(aq)H2O(l)+NaH2PO4(aq)Initial(mol)0.069(0.00150x)0Change(mol)0.00150x0.00150x+0.00150xAfter reaction(mol)(0

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