A solution of sodium acetate (NaCH3COO)has a pH of 9.73. The acid-dissociation constant for acetic acid...
A solution of sodium acetate (NaCH3COO)has a pH of 9.73. The acid-dissociation constant for acetic acid is 1.8×10−5 What is the molarity of the solution?
Express your answer to two significant figures and include the appropriate units.
Homework Answers
pKa = - log Ka = - log (1.8 * 10^-5) = 4.74
NaCH3COO is the salt of weak acid and strong base.
thus
salt hydrolysis occurs.
so Using
pH = 1/2 * [pKw + pKa + log C]
or
9.73 = 1/2 * [14 + 4.74 + log C ]
or
C = 5.2 M
thus
molarity of the solution = 5.2 M
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I need help with the solving this problem! Can someone please give a detailed explanation on...
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Calculate the pH of a solution made by adding 42 g of sodium acetate, NaCH3COO, to 22 g of acetic acid, CH3COOH, and dissolving in water to make 300. mL of solution. Hint given in feedback. The K, for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places. Answer: Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH3COO, to 22 g of acetic acid, CH3COOH, and dissolving in water...
please help!
Homework 5 on Chapter 16 Problem 16.81 13 of Review I Constants I Pen Part A A solution of sodium acetate (NaCHs COO) has a pH of 9.58. The acid-dissociation constant for acetic acid is 1.8 x 10.What is the molarity of the solution? Express your answer to two significant figures and include the appropriate units. A Units Value Molarity Request Answer Submit Provide Feedback Pearson I Contact Us Privacy PolicyI Permissions All rights reserved Terms
I need help with the solving this problem! Can someone please
give a detailed explanation on how to solve it?
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Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH3COO, to 22 g of acetic acid, CH COOH, and dissolving in water to make 500. mL of solution. Answer:
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