quation for the reaction that occurs when 0.090 mo

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
Chapter15: Solutions Of Acids And Bases
Section: Chapter Questions
Problem 15.82QE: Consider sodium acrylate, NaC3H3O2. Ka for acrylic acid (its conjugate acid) is 5.5 105. (a) Write...
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A buffer solution is made that is 0.352 M in H2S and 0.352 M in NaHS.
If Ka1 for H2S is 1.00 x 10 7, what is the pH of the buffer solution?
pH
Write the net ionic equation for the reaction that occurs when 0.090 mol HNO3 is added to 1.00 L of the buffer solution.
(Use the lowest possible coefficients. Omit states of matter. Use H30* instead of H*)
+
+
Transcribed Image Text:A buffer solution is made that is 0.352 M in H2S and 0.352 M in NaHS. If Ka1 for H2S is 1.00 x 10 7, what is the pH of the buffer solution? pH Write the net ionic equation for the reaction that occurs when 0.090 mol HNO3 is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter. Use H30* instead of H*) + +
A buffer solution is made that is 0.306 M in HCN and 0.306 M in KCN.
If K, for HCN is 4.00 x 10 10, what is the pH of the buffer solution?
pH
Write the net ionic equation for the reaction that occurs when 0.082 mol KOH is added to 1.00 L of the buffer solution.
(Use the lowest possible coefficients. Omit states of matter.)
Transcribed Image Text:A buffer solution is made that is 0.306 M in HCN and 0.306 M in KCN. If K, for HCN is 4.00 x 10 10, what is the pH of the buffer solution? pH Write the net ionic equation for the reaction that occurs when 0.082 mol KOH is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter.)
Expert Solution
Step 1

Given that:

[H2S] = 0.352 M
[NaHS] = 0.352 M
Ka = 1.00×10-7

Here H2S is a weak acid and NaHS is a salt of weak acid.

Step 2

pH of the buffer solution can be calculated by the Henderson-Hasselbalch equation.

pH=logKa+logSaltAcidpH=log1.00×107+log0.3520.352pH=7+0pH=7

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