Answered: What volumes of 0.200 MHCOOH and 2.00…

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter16: Reactions Between Acids And Bases

Section: Chapter Questions

Problem 16.41QE

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Question

What volumes of 0.200 MHCOOH and 2.00 MNaOHwould make 500. mL of a buffer with the pH of one made from475 mL of 0.200 Mbenzoic acid and 25 mL of 2.00 MNaOH?

Expert Solution

Step 1

This problem is solved by using Henderson equating for calculation of pH of buffer.

$p H = p K_{a} - \log [\frac{(w e a k a c i d)}{(c o j u g a t e b a s e)}]$

Step 2

Using Henderson's equation:

$p H = p K_{a} - \log [\frac{(w e a k a c i d)}{(c o j u g a t e b a s e)}]$

$[H C O O H] = 0.2 M [N a O H] = 2.0 M [B e n z o i c a c i d] = 0.2 M$

Now , we hav to calculate number of moles of benzoic acid:

= molarity* volume

$= 2 \times 0.475 = 0.095 m o l$

Step 3

Also, number of moles of NaOH:

$= 2 \times 0.025 = 0.05 m o l$

$C_{6} H_{5} C O O H + N a O H \to C_{6} H 5 C O O N a + H_{2} O$

$C_{6} H_{5} C O O H$	NaOH	$C_{6} H_{5} C O O N a$	$H_{2} O$
0.095	0.05	0	0
0.095-0.05	0	0.05

$\begin{array}{rcl} p H & = & p K_{a} + \log \frac{B a s e}{A c i d} \\ = & 4.2 + \log \frac{0.05}{0.045} \\ = & 4.245 \end{array}$

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