Chapter 17, Problem 110IL

### 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

# What mass of Na3PO4 must be added to 80.0 mL of 0.200 M HCl to obtain a buffer with a pH of 7.75?

Interpretation Introduction

Interpretation:

The mass of Na3PO4 has to be calcualted which is added to 80 mL of 0.200 M HCl to give the buffer solution of pH value equals to 7.75.

Concept introduction:

The reaction between strong base Na3PO4 and  strong acid HCl gives the product Na2HPO4  which is the conjugate acid of Na3PO4. The Na2HPO4 further react with the remaining  HCl and forms NaH2PO4.

The aqueous solution of HCl gives H3O+ which react with Na3PO4 and Na2HPO4.

HCl(aq)+H2O(l)H3O+(aq)+Cl(aq)

The equilibrium between Na3PO4 and H3O+ is written as follows;

Na3PO4(aq)+H3O+(aq)Na2HPO4(aq)+H2O(l)

The equilibrium between Na2HPO4 and H3O+ is written as follows;

Na2HPO4(aq)+H3O+(aq)NaH2PO4(aq)+H2O(l)

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,

pOH=pKb+log[conjugateacid][base] (1)

Explanation

The calculation for amount of Na3PO4 is done by using Henderson-Hasselbalch equation.

Given:

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

The value of Kb for Na3PO4 is 2.8Ã—10âˆ’2.

The value of Kb for Na2HPO4 is 1.6Ã—10âˆ’7.

Negative logarithm of the Kb value gives the pKb value of the acid

Therefore, pKb value for Na3PO4 is 1.55 and for Na2HPO4 pKb value is 6.80.

The value of pH for the solution is 7.75.

The pOH value of the solution is calculated.

pOHÂ =Â 14âˆ’7.75=6.25

Therefore, pOH value of the solution is 6.25.

Let xÂ mol of Na3PO4 is added to the HCl solution.

The concentration of HCl solution is 0.200Â molâ‹…Lâˆ’1.

Volume of HCl solution is 0.080Â L.

The calculation of moles is done by using the expression,

Numberâ€‰ofÂ moles=concentration(molâ‹…Lâˆ’1)â‹…volume(L)

The ICE table (1) for the reaction between HCl and Na3PO4 is given below,

EquationNa3PO4(aq)+H3O+(aq)â†’H2O(l)+Na2HPO4(aq)Initial(mol)x(0.016)0Change(mol)âˆ’xâˆ’x+xAfterÂ reaction(mol)0(0.016âˆ’x)x

The generated moles of Na2HPO4 that is xÂ mol will further react with remainingÂ  (0.016âˆ’x)mol of H3O+.

The ICE table (2) for the reaction between H3O+ and Na2HPO4 is given below,

EquationNa2HPO4(aq)+H3O+(aq)â†’H2O(l)+NaH2PO4(aq)Initial(mol)x(0

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