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tamaranim1 [39]
3 years ago
6

What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu-

tion having a pH of 3.74? Ka = 0.00018 for HCOOH.
Answer in units of g.
Chemistry
1 answer:
Sergio [31]3 years ago
4 0

Answer:

We need 4.28 grams of sodium formate

Explanation:

<u>Step 1:</u> Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

<u>Step 2:</u> Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

<u>Step 3:</u> Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

<u>Step 4:</u> Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate

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