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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,

$K_{w} = K_{a} 脳 K_{b}$

02

Step 2:Calculate the PH of 0.050M sodium bicarbonate

The chemical equation can be given as,

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

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

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

The pH of weak base is calculated using the equation,

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

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

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Most popular questions from this chapter

Why doesn鈥檛 water produce $10^{- 7} M H^{+}$ , andwhen

is added?

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If, instead, mmol of strong base were generated, by how much would the pH of the original buffer rise?

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