Answered: what buffer component ratio base/acid…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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what buffer component ratio base/acid would be needed to make a pH 11.75 buffer using a mixture C5H10NH2

Expert Solution

Step 1

Required pH of buffer is 11.75

Given component is C₅H₁₀NH₂⁺.

The given component is a conjugated acid for C₅H₁₀NH . So, in order to make a buffer a weak base C₅H₁₀NH is to be mixed with C₅H₁₀NH₂⁺.

The weak base will be C₅H₁₀NH.

pH of a buffer is given Hassel Bach - Henderson's equation as follows:

$p H = p K_{a} + \log \frac{[B a s e]}{[A c i d]}$

we know that, pH+pOH =14

pOH=14-11.75

pOH=2.25

So, for a weak base the pOH of a buffer is given by:

$p O H = p K_{b} + \log \frac{[A c i d]}{[B a s e]}$

Dissociation constant of weak base (C₅H₁₀NH) is = 1.3 x 10^-3

pK_b is given by:

$p K_{b} = - \log K_{b} p K_{b} = - \log (1.3 \times 10^{- 3}) p K_{b} = - (- 2.886) p K_{b} = 2.886$

Now substituting the values in Hassel Bach - Henderson's equation :

$p O H = p K_{b} + \log \frac{[A c i d]}{[B a s e]} \log \frac{[A c i d]}{[B a s e]} = p O H - p K_{b} \log \frac{[A c i d]}{[B a s e]} = 2.25 - 2.886 \log \frac{[A c i d]}{[B a s e]} = - 0.636 \frac{[A c i d]}{[B a s e]} = {(- 0.636)}^{10} \frac{[A c i d]}{[B a s e]} = 0.2312 \frac{[B a s e]}{[A c i d]} = \frac{1}{0.2312} \frac{[B a s e]}{[A c i d]} = 4.3252$

Therefore, the buffer component ratio for Base/Acid needed to make a pH of 11.75 is 4.325

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