Calculate the pH of the solution that results from each of the following mixture:
160.0 mL of 0.25 M HF with 225.0 mL of 0.31 M NaF
Ka of HF = 6.6*10^-4
Concentration after mixing = mol of component / (total volume)
M(F-) after mixing = M(F-*)V(F-)/(total volume)
M(F-) after mixing = 0.31 M*225.0 mL/(225.0+160.0)mL
M(F-) after mixing = 0.1812 M
Concentration after mixing = mol of component / (total volume)
M(HF) after mixing = M(HF*)V(HF)/(total volume)
M(HF) after mixing = 0.25 M*160.0 mL/(160.0+225.0)mL
M(HF) after mixing = 0.1039 M
Ka = 6.6*10^-4
pKa = - log (Ka)
= - log(6.6*10^-4)
= 3.1805
we have below equation to be used:
This is Henderson–Hasselbalch equation
pH = pKa + log {[conjugate base]/[acid]}
= 3.1805+ log {0.1812/0.1039}
= 3.4219
Answer: 3.42
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