How many grams of solid NaC2H3O2•3H2O (MW 136.09 g/mol) must be added to 0.300 L of a 0.50-M acetic acid (HC2H3O2) solution to give a buffer with a pH of 5.00? Assume a negligible change in volume as the solid is added. The pKa for acetic acid is 4.76. (Hint: What is the ratio of conjugate base (C2H3O2-) to acid (HC2H3O2) needed to make this buffer?)

We have to add 35.48 grams of NaC2H3O2•3H2O

Explanation:

Step 1: data given

Volume of a 0.50 M CH3COOH = 0.300 L

pH = 5.0

The pKa for acetic acid is 4.76

Step 2: Calculate [CH3COO-]

pH = pKa + log[CH3COO-]/[CH3COOH]

5.0 = 4.76 + log[CH3COO-]/[CH3COOH]

log[CH3COO-]/[CH3COOH] = 0.24

[CH3COO-]/[CH3COOH] = 10^0.24

[CH3COO-]/[CH3COOH] = 1.738

[CH3COO-]/0.50 M  = 1.738

[CH3COO-] = 0.50 * 1.738

[CH3COO-] = 0.869 M

Step 3: Calculate moles CH3COO-

Moles = molarity * volume

Moles CH3COO- = 0.869 M * 0.300 L

Moles CH3COO- = 0.2607 moles

Step 4: Calculate moles CH3COONa

For 1 mol CH3COONa we have 1 mol CH3COO-

For 0.2607 moles CH3COO- we need 0.2607 moles CH3COONa

Step 5: Calculate mass CH3COONa

Mass CH3COONa = moles * molar mass

Mass CH3COONa = 0.2607 moles * 136.09 g/mol

Mass CH3COONa = 35.48 grams

We have to add 35.48 grams of NaC2H3O2•3H2O

the answer for sodium is 0.186

the answer for rubidium is 0.251

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