PH of 0-05 M CH 3 COO 2 Ca PK a -4-74 isA- 8-87B- 7C- 1-3D- 8-72

The correct option is A 8.87[CH_3COO^–] = 2 × 0.05 = 0.1 pH = 1/2 [pK_w + pK_a + log C] = 1/2 [14 + 4.74 + log0.1] = 1/2 [14 + 4.74 1] = 8.87 ...

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Byju's Answer

Standard XII

Chemistry

Revisiting Tautomerism

pH of 0.05 M ...

Question

pH of $0.05 M (C H_{3} C O O)_{2} C a (p K_{a} = 4.74)$ is –

8.72

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8.87

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1.3

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Solution

The correct option is B 8.87
$[C H_{3} C O O^{-}] = 2 \times 0.05 = 0.1$
$p H = \frac{1}{2} [p K_{w} + p K_{a} + l o g C] = \frac{1}{2} [14 + 4.74 + l o g 0.1] = \frac{1}{2} [14 + 4.74 - 1] = 8.87$

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pH of 0.05M(CH3COO)2Ca(pKa=4.74) is –

The correct option is B 8.87[CH3COO–]=2×0.05=0.1 pH=12[pKw+pKa+log C]=12[14+4.74+log0.1]=12[14+4.74−1]=8.87

pH of $0.05 M (C H_{3} C O O)_{2} C a (p K_{a} = 4.74)$ is –

The correct option is B 8.87
$[C H_{3} C O O^{-}] = 2 \times 0.05 = 0.1$
$p H = \frac{1}{2} [p K_{w} + p K_{a} + l o g C] = \frac{1}{2} [14 + 4.74 + l o g 0.1] = \frac{1}{2} [14 + 4.74 - 1] = 8.87$