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Chapter 9: Q9FE (page 207)

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

Short Answer

Expert verified

The of Sodium butanoate is 8.76.

Step by step solution

01

Th base dissociation constant

The K_b base dissociation constant is a measure of basicity or the strength of the base in general.

The value ofis:

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

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 butanoate

The chemical equation can be given as,

${CH}_{3} {CH}_{2} {CH}_{2} {CO}_{2}^{-} H_{2} O 鈬� {CH}_{3} {CH}_{2} {CH}_{2} {CO}_{2} H + {OH}^{-}$

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

$K_{b} = \frac{K_{w}}{K_{a}} = 6.58 脳 10^{- 10}$

$FromICEtable : \frac{x^{2}}{0.050 - x} = K_{b} x = \sqrt{(0.050) (6.58 脳 10^{- 10}) x} = 5.74 脳 10^{- 6} M$

The pH of weak base is calculated using the equation,

$pH = [H^{+}] = \frac{K_{w}}{[{OH}^{-}]} pH = - \log [H^{+}] pH = - \log [5.74 脳 10^{- 6} M] pH = 8.76$

The pH of 0.050M Sodium butanoate is 8.76.

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Most popular questions from this chapter

${Cr}^{3 +}$ is acidic by virtue of the hydrolysis reaction ${Cr}^{3 +} H_{2} O \overset{K_{aI}}{鈬�} Cr {(OH)}^{2 +} + H^{+}$ . [Further reactions produce $Cr {(OH)}_{2,}^{+} Cr {(OH)}_{3}$ and.] Find the value role="math" localid="1654864894349" $K_{aI}$ of in Figure 6-9. Considering only the $K_{aI}$ reaction, find pH value ofrole="math" localid="1654865110221" $0.010 MCr {({CIO}_{4})}_{3}$ What fraction of chromium is in the formrole="math" localid="1654865163335" $Cr {(OH)}^{2 +}$ ?

Find $[OCI 2] / [HOCI]$ if pH is raised by one unit to 7.20.

Write the chemical reaction whose equilibrium constant is

(a) $K_{a}$ for benzoic acid, $C_{6} H_{5} {CO}_{2} H$

(b) $K_{b}$ for benzoate ion, $C_{6} H_{5} {CO}_{2}^{2 -}$

(c) $K_{b}$ for aniline, $C_{6} H_{5} {NH}_{2}$

(d) $K_{a}$ for anilinium ion, $C_{6} H_{5} {NH}_{3}^{+}$

Calculate how many milliliters of $0.626 MKOH$ should be added to $5.00 g$ of MOBS (Table 9-2) to give a pH of 7.40.

Why does buffer capacity increase as a solution becomes very acidic (pH 鈮� 1) or very basic (pH 鈮� 13)?

See all solutions

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