   # Use the Henderson-Hasselbalch equation to calculate the pH of 1.00 L of a buffer solution containing 15.0 g of NaHCO 3 and 18.0 g of Na 2 CO 3 . (Consider this buffer as a solution of the weak acid HCO 3 − and its conjugate base, CO 3 2− .) ### 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 17.2, Problem 2CYU
Textbook Problem
47 views

## Use the Henderson-Hasselbalch equation to calculate the pH of 1.00 L of a buffer solution containing 15.0 g of NaHCO3 and 18.0 g of Na2CO3. (Consider this buffer as a solution of the weak acid HCO3− and its conjugate base, CO32−.)

Interpretation Introduction

Interpretation:

The value of pH for the solution containing 15.0g of sodium bicarbonate and 18.0g of sodium carbonate were dissolved in 1litre of water is to be calculated by using Henderson-Hasselbalch equation.

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, is strength of the acid which can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

### Explanation of Solution

The equilibrium between HCO3 and its conjugate base CO32 is written as follows;

HCO3(aq)+H2O(l)H3O+(aq)+CO32(aq)(acid)(conjugatebase)

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

Given:

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

The value of Ka for hydrogen carbonate ion is 4.8×1011.

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

pKa=log(Ka)=log(4.8×1011)=10.31

Therefore, pKa value for the hydrogen carbonate ion is 10.31.

The 15.0g of sodium bicarbonate (NaHCO3) is dissolved in 1litre of H2O.

Molar mass of sodium bicarbonate (NaHCO3) is 84.0gmol1.

The 18.0g of sodium carbonate (Na2CO3) is dissolved in 1litre of H2O.

Molar mass of sodium carbonate is 105.98gmol1.

The concentration of sodium bicarbonate is calculated as follows;

Molarity=Numberofmoles1L of solvent (2)

Numberofmoles are calculated by using expression.

Numberofmoles=weightmolarmass

Substitute 15.0g for weight and 84.0gmol1 for molarmass for sodium bicarbonate (NaHCO3)

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
For each atomic symbol, give the name of the element. a Be b Ag c Si d C

General Chemistry - Standalone book (MindTap Course List)

Most codons, specify a(n) ___ . a. protein b. polypeptide c. amino acid d. mRNA

Biology: The Unity and Diversity of Life (MindTap Course List)

Fat-soluble vitamins are mostly absorbed into a. the lymph. b. the blood. c. the extracellular fluid. d. b and ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

How are the heaviest elements (uranium or gold) thought to be formed?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

Will the Sun pass through the instability strip? Why or why not?

Foundations of Astronomy (MindTap Course List)

A sound wave moves down a cylinder as in Figure 16.17. Show that the pressure variation of the wave is describe...

Physics for Scientists and Engineers, Technology Update (No access codes included) 