3.A solution was prepared by adding 4.95 g sodium acetate (NaC;H;O2) to 250.0 mL of 0.150 Macetic acid (K = 1.75 x 10). Assume negligible volume change.a)What is the pH of the resulting solution?PH: pk2 + logCH3 02)moles saium4.9590.06 mol%3D2cent%3D82gmuH: -log (1.73x 10) +log0.24moles 2cete250 x0.15IL :003751000mePH : 4 1.96O.06(N2 G Hs0): 0.24m0.230 LO.0375: CCHS COOH): 0 15 mAssume 125.0 mL of the solution prepared above was put into a beaker and 82 mg ofNaOH was added to it. What is the pH of this new mixture?b)

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Answered: Assume 125.0 mL of the solution…

Assume 125.0 mL of the solution prepared above was put into a beaker and 82 mg of NaOH was added to it. What is the pH of this new mixture? b)

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter14: Equilibria In Acid-base Solutions

Section: Chapter Questions

Problem 33QAP: A buffer is prepared in which the ratio [ H2PO4 ]/[ HPO42 ]is 3.0. (a) What is the pH of this...

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

3.
A solution was prepared by adding 4.95 g sodium acetate (NaC;H;O2) to 250.0 mL of 0.150 M
acetic acid (K = 1.75 x 10). Assume negligible volume change.
a)
What is the pH of the resulting solution?
PH: pk2 + log
CH3 02)
moles saium
4.959
0.06 mol
%3D
2cent
%3D
82gmu
H: -log (1.73x 10) +
log
0.24
moles 2cete
250 x
0.15
IL :00375
1000me
PH : 4 1.96
O.06
(N2 G Hs0): 0.24m
0.230 L
O.0375
: CCHS COOH): 0 15 m
Assume 125.0 mL of the solution prepared above was put into a beaker and 82 mg of
NaOH was added to it. What is the pH of this new mixture?
b)

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