Chapter 17, Problem 24PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# You dissolve 0.425 g of NaOH in 2.00 L of a buffer solution that has [H2PO4−| = [HPO42−] = 0.132 M. What is the pH of the solution before adding NaOH? After adding NaOH?

Interpretation Introduction

Interpretation:

For the buffer solution which contains H2PO4 and HPO42 in equal amount of concentration the pH value is to be calculated before and after addition of 0.425 g NaOH.

Concept introduction:

The Henderson-Hasselbalch equation relates pH of a buffer with pKa of acid, concentration of conjugate base and concentration of acid. The expression is written as,

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

This equation shows that pH of buffer solution is controlled by two major factors. First, Strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

Explanation

The calculation of pH is done by using Henderson-Hasselbalch equation.

Given:

Refer to table no. 16.2 in the textbook for the value of Ka.

The value of Ka for H2PO42âˆ’ is 6.2Ã—10âˆ’8.

Negative logarithm of the Ka value gives the pKa value of the acid.

pKa=âˆ’log(Ka)=âˆ’log(6.2Ã—10âˆ’8)=7.20

Therefore, pKa value for the H2PO4âˆ’ is 7.20.

The concentration of H2PO4âˆ’ is 0.132Â M.

The concentration of HPO42âˆ’ is 0.132Â M.

The volume of the buffer solution is 2Â L.

The value of pH is calculated by using equation (1).

pH=pKa+log[conjugateâ€‰base][acid]

Substitute 7.20 for pKa, 0.132Â  for [acid], 0.132Â  for

[conjugateâ€‰base].

pH=7.20+log(0.132Â )(0.132Â )=7.20+log(1.0)=7.20+0=7.20

Therefore, pH value of the buffer solution before addition of NaOHÂ  is 7.20.

Calculation of pH after addition of NaOH is calculated below.

The added mass of NaOH is 0.425Â g.

The molar mass of NaOH is 40Â gâ‹…molâˆ’1.

For NaOH Number of moles are calculated by using expression.

NumberÂ ofÂ moles=weightmolarÂ mass

Substitute 0.425Â g for weight and 40Â gâ‹…molâˆ’1 for molarâ€‰mass.

NumberÂ ofÂ moles=0.425Â g40Â gâ‹…molâˆ’1=0.010Â mol

Therefore number of moles of NaOH added are 0.010Â mol at the equilibrium between H2PO4âˆ’ and HPO42âˆ’.

As, NaOH is a strong base it ionize completely into its ions in aqueous solution.

NaOH(aq)â†’Na+(aq)+Â OHâˆ’(aq)

The hydroxide ions will react with H2PO4âˆ’ and affects the equilibrium conditions.

H2PO4âˆ’(aq)+Â OHâˆ’(aq)â†’H2O(l)Â +Â HPO42âˆ’(aq)

Table gives the relationship of the concentration for the reaction between OHâˆ’, H2PO4âˆ’ and HPO42âˆ’.

Â Â Â Â Â Â Â Â Â Â Â OHâˆ’(fromÂ addedÂ NaOH)Â Â Â Â Â H2PO4âˆ’(fromÂ buffer)Â Â Â Â Â HPO42âˆ’(fromÂ buffer)initial(mol)0.0100.2640.264change(mol)âˆ’0.010âˆ’0

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