The dissociation constant of a substituted benzoic acid at 25- C is 1-0 - 10-4- The pH of a 0-01 M solution of its sodium salt is

K_a(C_6H_5COOH)=1 × 10^ 4 pH of 0.01 M C_6H_5COONa 0.01 (1 h)C_6H_5COO^ + 0.01 hH_2O⇌0.01 hC_6H_5COOH + OH^ K_b = K_w/K_a = 0.01 h^2/1 h 10^ 14/10^ 4 = 10^ 2 ...

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Standard XII

Chemistry

Ostwald Dilution Law Mathematical Interpretation

The dissociat...

Question

The dissociation constant of a substituted benzoic acid at $25^{\circ} C$ is $1.0 \times 10^{- 4}$ . The pH of a 0.01 M solution of its sodium salt is ___

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Solution

$K_{a} (C_{6} H_{5} C O O H) = 1 \times 10^{- 4}$
pH of 0.01 M $C_{6} H_{5} C O O N a$
$C_{6} H_{5} C O O^{-} 0.01 (1 - h) + H_{2} O 0.01 h ⇌ C_{6} H_{5} C O O H 0.01 h + O H^{-}$
$K_{b} = \frac{K_{w}}{K_{a}} = \frac{0.01 h^{2}}{1 - h}$
$\frac{10^{- 14}}{10^{- 4}} = \frac{10^{- 2} h^{2}}{1 - h}, (1 - h \approx 1)$
$[O H^{-}] = 0.01 h = 0.01 \times 10^{- 4} = 10^{- 6}$
$[H^{+}] = 10^{- 8}$
$p H = 8$

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The dissociation constant of a substituted benzoic acid at 25∘C is 1.0×10−4. The pH of a 0.01 M solution of its sodium salt is ___

Ka(C6H5COOH)=1×10−4 pH of 0.01 M C6H5COONa C6H5COO−0.01(1−h)+H2O0.01 h⇌C6H5COOH0.01 h+OH− Kb=KwKa=0.01 h21−h 10−1410−4=10−2 h21−h, (1−h≈1) [OH−]=0.01 h=0.01×10−4=10−6 [H+]=10−8 pH=8

The dissociation constant of a substituted benzoic acid at $25^{\circ} C$ is $1.0 \times 10^{- 4}$ . The pH of a 0.01 M solution of its sodium salt is ___