Suppose you have a 250 mL buffer solution containing 0.400 M nitrous acid and 0.
ID: 815699 • Letter: S
Question
Suppose you have a 250 mL buffer solution containing 0.400 M nitrous acid and 0.400 M sodium nitrite. You then add 50.0 mL of a 0.100 M NaOH solution. What is the final pH of the system assuming that the buffer concentration were at equilibrium before the base was added. What would the pH be if 50.0 mL of 0.100 M NaOH was added to 250 mL of pure water without a buffer.
New Buffer pH = 2.98; pH without buffer = 10.3
New Buffer pH = 3.71; pH without buffer = 12.6
New Buffer pH = 3.39; pH without buffer = 12.2
New Buffer pH = 3.54; pH without buffer = 11.7
New Buffer pH = 4.13; pH without buffer = 13.1
New Buffer pH = 2.98; pH without buffer = 10.3
New Buffer pH = 3.71; pH without buffer = 12.6
New Buffer pH = 3.39; pH without buffer = 12.2
New Buffer pH = 3.54; pH without buffer = 11.7
New Buffer pH = 4.13; pH without buffer = 13.1
Explanation / Answer
New Buffer pH:
pH = pKa - Log [salt]x[Base]/[acid]
pH = 3.14 - Log [0.4M][0.1M]/[0.4M]
pH = 4.14
IF:
Without Buffer pH:
pH = 14 - pOH
pH = 14 - Log [OH-]
AND IF:
[OH-]=?
V1C1 = V2C2
50 mL x 0.100M = 300 mL x C2
C2=(50 mL x 0.100M)/300 mL
C2=0.01667M
THEN:
pH = 14 + Log [OH-]
pH = 14 + Log [0.01667M]
pH = 14 + Log [0.01667M]
pH = 12.22
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