   Chapter 17, Problem 81GQ

Chapter
Section
Textbook Problem

Calculate the hydronium ion concentration and the pH of the solution that results when 50.0 mL of 0.40 M NH3 is mixed with 25.0 mL of 0.20 M HCl.

Interpretation Introduction

Interpretation:

The hydronium ion concentration and pH for the resultant solution has to be calculated when 50 mL of 0.40 M NH3 is mixed with 25 mL of 0.20 M NaOH.

Concept introduction:

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

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

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

The pH value is calculated by expression, pH + pOH = 14.

Explanation

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

Given:

Refer to table 16.2 in the textbook for the value of Kb.

The value of Kb for NH3 is 1.8×105.

The pKb value is calculated as follows;

pKb=log(Kb)

Substitute, 1.8×105 for Kb.

pKb=log(1.8×105)=4.74

Therefore, pKb value of NH3 is 4.74.

The initial concentration of NH3 is 0.40 molL1.

The volume of buffer solution is 50 mL.

Unit conversion of 50 mL into L.

(50 mL)(1 L1000 mL)=0.050 L

Therefore, volume of the solution is 0.050 L.

The concentration of HCl is 0.20 M, and volume of HCl added is 25 mL.

Unit conversion of 25 mL into L.

(25 mL)(1 L1000 mL)=0.025 L

Therefore, volume of the HCl solution is 0.025 L.

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

HCl(aq) + H2O(l)H3O+(aq)+ Cl(aq)

The H3O+ ions will react with ammonia.

NH3(aq)+ H3O+(aq)H2O(l) + NH4+(aq)

Table gives the relationship of the concentration for the reaction between H3O+, NH3 and NH4+.

H3O+(from added HCl)NH3             NH4+  initial(mol)0.0050.020.0change(mol)0.0050.005+0.005after reaction(mol)00.0150.005

The concentration of ammonia and ammonium ion after the reaction is calculated by using equation (2);

Molarity=Number of molesvolume of solvent (2)

The total volume of the solution will increase by 0.025 L after reaction.

Therefore, total volume of the solution is calculated as,

0.050 L+ 0.025 L = 0.075 L

Total volume of the solution is  0.075 L.

Substitute, 0.015 mol for Number of moles and  0.075 L for  volume of solvent to calculate the concentration of ammonia.

Molarity=0.015 mol0.075 L=0.2 molL1

Therefore, concentration of ammonia ion left after reaction is 0.2 molL1.

Substitute, 0.005 mol for Number of moles and  0.075 L for  volume of solvent to calculate the concentration of ammonia in equation (2)

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