You dissolve 0.356 g of NaOH in 2.00 L of a buffer solution that originally had
ID: 103576 • Letter: Y
Question
You dissolve 0.356 g of NaOH in 2.00 L of a buffer solution that originally had [H2PO4-] = [HPO42-]= 0.132 M. Look up the Ka value in your text.
a. What is the pH of the buffer solution before addition of the sodium hydroxide?
pH =
b. What is the concentration of dihydrogen phosphate after addition of the sodium hydroxide?
[H2PO4-] = M
c. What is the concentration of the hydrogen phosphate after addition of the sodium hydroxide?
[HPO42-] = M
d. What is the pH of the buffer solution after addition of the sodium hydroxide?
pH =
Explanation / Answer
a.
initially
pH = pKa + log(HPO4-2/H2PO4-)
pH = 7.21 + log(1)
pH = 7.21
b and c)
after addition
mol of NaOH = mass/W = 0.356/40 = 0.0089
mol of H2PO4- = 0.132*2 - 0.0089 = 0.2551
[H2PO4-2] = mol/V = 0.2551/2 =0.12755 M
mol of HPO4-2 = 0.132*2 + 0.0089 = 0.2729
[H2PO4-2] = mol/V = 0.2729/2 =0.13645 M
d
pH = pKa + log(HPO4-2/H2PO4-)
pH = 7.21 + log(0.2729/0.2551)
pH = 7.23929
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