Calculate the ratio of CH 3 N H 2 to CH 3 N H 3- Cl -required to create a buffer with pH -10-14K b of CH 3 NH 2-4-4 - 10-4take log 2-27-0-356- log 0-313-0-504A- 0-5B- 0-31C- None of the aboveD- 1-05

The correct option is B 0.31Given, nbsp;K_b of CH_3NH_2 = 4.4 × 10^ 4 , pH=10.14 We need the K_a of the methylammonium ion: nbsp; We know at 25^oC K_a× K ...

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Standard XII

Chemistry

Basic Buffer Action

Calculate the...

Question

Calculate the ratio of $C H_{3} N H_{2}$ to $C H_{3} N H_{3}^{+} C l^{-}$ required to create a buffer with pH = 10.14
$K_{b} o f C H_{3} N H_{2} = 4.4 \times 10^{- 4}$
take $l o g 2.27 = 0.356, l o g 0.313 = - 0.504$

0.5

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1.05

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0.31

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None of the above

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Solution

The correct option is C 0.31
Given, $K_{b} o f C H_{3} N H_{2} = 4.4 \times 10^{- 4}, p H = 10.14$
We need the $K_{a}$ of the methylammonium ion:
We know at $25^{o} C K_{a} \times K_{b} = 10^{- 14}$
$K_{a} for C H_{3} N H_{3}^{+} C l^{-} = 10^{- 14} / 4.4 \times 10^{- 4} = 2.27 \times 10^{- 11}$
Using the Henderson-Hasselbalch equation:
$p H = p K_{a} + l o g [\frac{[conjugate base]}{[acid]}]$
Substitute and solve for the base/acid ratio:
$10.14 = - l o g 2.27 \times 10^{- 11} + l o g [\frac{[conjugate base]}{[acid]}]$
$10.14 = 11 - 0.356 + l o g [\frac{[conjugate base]}{[acid]}]$
$10.14 = 10.644 + l o g [\frac{[conjugate base]}{[acid]}]$
$l o g [\frac{[conjugate base]}{[acid]}] = - 0.504$
$\frac{[conjugate base]}{[acid]} = a n t i l o g (- 0.504)$
$\frac{[conjugate base]}{[acid]} = 0.313$

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Basic Buffer Action

Standard XII Chemistry

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Calculate the ratio of CH3NH2 to CH3NH+3Cl− required to create a buffer with pH = 10.14 Kb of CH3NH2=4.4×10−4 take log 2.27=0.356, log 0.313=−0.504

Calculate the ratio of $C H_{3} N H_{2}$ to $C H_{3} N H_{3}^{+} C l^{-}$ required to create a buffer with pH = 10.14
$K_{b} o f C H_{3} N H_{2} = 4.4 \times 10^{- 4}$
take $l o g 2.27 = 0.356, l o g 0.313 = - 0.504$