PH of 0-5 M Ba CN 2 solution pK b - of -CN -9-30 is-A- 8-35B- 3-35C- 9-35D- 9-50

The correct option is C 9.35 [CN^ ] = 2 × 0.5 = 1.0 M pH = 1/2 (pK_w + pK_a + log[CN^ ]) = 1/2 [14 + (14 pK_b) + log 1] = 1/2[28 9.30] = 9.35 ...

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Byju's Answer

Standard XII

Chemistry

Henderson Hassleback Equation

pH of 0.5 M B...

Question

pH of 0.5 M $B a (C N)_{2}$ solution $(p K_{b} o f C N^{-} = 9.30)$ is :

8.35

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3.35

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9.35

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9.50

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Solution

The correct option is C
9.35

$[C N^{-}] = 2 \times 0.5 = 1.0 M p H = \frac{1}{2} (p K_{w} + p K_{a} + l o g [C N^{-}]) = \frac{1}{2} [14 + (14 - p K_{b}) + l o g 1] = \frac{1}{2} [28 - 9.30] = 9.35$

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pH of 0.5 M Ba(CN)2 solution (pKb of CN−=9.30) is :

The correct option is C 9.35 [CN−]=2×0.5=1.0MpH=12(pKw+pKa+log[CN−])=12[14+(14−pKb)+log 1]=12[28−9.30]=9.35

pH of 0.5 M $B a (C N)_{2}$ solution $(p K_{b} o f C N^{-} = 9.30)$ is :

The correct option is C
9.35

$[C N^{-}] = 2 \times 0.5 = 1.0 M p H = \frac{1}{2} (p K_{w} + p K_{a} + l o g [C N^{-}]) = \frac{1}{2} [14 + (14 - p K_{b}) + l o g 1] = \frac{1}{2} [28 - 9.30] = 9.35$