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Chapter 15: Problem 157

Calculate the \(\mathrm{pH}\) of a \(0.20 \mathrm{M} \mathrm{NaHCO}_{3}\) solution. (Hint: As an approximation, calculate hydrolysis and ionization separately first, followed by partial neutralization.)

Short Answer

Expert verified
The pH of the 0.20 M NaHCO3 solution is 9.83.

Step by step solution

01

Ionization of NaHCO3

In water, the ionic compound NaHCO3 dissociates completely into Na+ and HCO3- ions. Thus, the concentration of HCO3- is 0.20 M.
02

Calculate Hydrolysis

The hydrolysis reaction of HCO3- in water can be given as \(HCO_{3}^- + H_{2}O \rightleftharpoons H_{2}CO_{3} + OH^-\). The hydrolysis constant can be expressed as \(K_h = K_w / K_a\), where \(K_w\) is the ion product of water (1.0 x 10^-14 at 25 degrees Celsius), and \(K_a\) is the acid dissociation constant of \(H_2CO_{3}\) (4.3 x 10^-7). Thus, \(K_h = (1.0 x 10^-14) / (4.3 x 10^-7) = 2.3 x 10^-8 \). Using this, the hydroxide ion concentration (\([OH^-]\)) can be determined using the expression \( [OH^-] = sqrt([HCO_{3}^-] x Kh) = sqrt(0.20 x 2.3 x 10^-8) = 6.8 x 10^-5 M \).
03

Calculate pH

With the hydroxide ion concentration, the pOH can be calculated by using the formula \( pOH = -log[OH^-] = -log(6.8 x 10^-5) = 4.17 \). Since pH + pOH = pKw (which is 14 at 25 degrees Celsius), therefore, the pH of the solution would be pH = 14 - pOH = 14 - 4.17 = 9.83.

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Most popular questions from this chapter

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