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Chapter 9: QFE (page 207)

Find the pH of 0.050 M sodium butanoate (the sodium salt ofbutanoic acid, also called butyric acid).

Short Answer

Expert verified

The pH of 0.050M Sodium butanoate is=8.76

Step by step solution

01

Define base dissociation constant.

The K b base dissociation constant is a measure of basicity, or the strength of the base in general.

The value of $K_{b} K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}$

Where $K_{b}$ is called as base dissociation constant.

The value of $K_{w}$ is calculated by the formula,

$K_{w} = K_{a} \times K_{b}$

02

Step 2:Calculate the PH of 0.050M sodium bicarbonate

The chemical equation can be given as,

${CH}_{3} {CH}_{2} {CH}_{2} {CO}_{2}^{-} + H_{2} O ֏ {CH}_{3} {CH}_{2} {CH}_{2} {CO}_{2} H + {OH}^{-} 8 F - x x + x$

The value of $K_{b}$ is calculated from the value of $K_{w}$ and $K_{a}$

$K_{b} = \frac{K_{w}}{K_{a}} = 6.58 \times 10^{- 10} \frac{x^{2}}{F - x} = \frac{x^{2}}{0.050 - x} = K_{b} x = \sqrt{(0.050) (6.58 \times 10^{- 10}) x} = 5 . 7_{4} \times 10^{- 6} M$

The pH of weak base is calculated using the equation,

$pH = [H^{+}] = \frac{K_{w}}{[OH]} pH = - \log [H^{+}] pH = - \log [5 . 7_{4} \times 10^{- 6} M] pH = 8.76$

The pH of $0.050 M$ Sodium butanoate $= 8.76$

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Most popular questions from this chapter

Consider a reaction mixture containing 100.0mLof 0.100mLborate buffer at $pH = {pK}_{3}$ = 9.24 . At $pH = {pK}_{a}$ , we know that $[H_{3} {BO}_{3}] = [H_{2} {BO}_{3}] = 0.0500 M$ . Suppose that a chemical reaction whose pHwe wish to control will be generating acid. To avoid changing the pH very much we do not want to generate more acid than would use up half of the $[H_{2} {BO}_{3}]$ .How many moles of acid could be generated without using up more than half of the $[H_{2} {BO}_{3}]$ ?What would bepH?

How many milliliters of $0.246 {MHNO}_{3}$ should be added to $213 mL$ of $0.00666 M 2,$ -bipyridine to give a pH of $4.19$ ?

A $0.0450 M$ solution of benzoic acid has a pH of $2.78$ . Calculate ${pK}_{a}$ for this acid

What is the pH of a solution prepared by dissolving $1.23 g$ of

2-nitrophenol (FM 139.11) in 0.250 L?

Using activity coefficients correctly, find the pH of $1.0 \times 10^{- 2} MNaOH$ .

See all solutions

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