First, calculate the moles of HF and NaOH. The initial moles of HF are given by:
moles of HF=0.0380L×0.55M=0.0209mol
The moles of NaOH added are:
moles of NaOH=0.0220L×0.55M=0.0121mol
Since NaOH reacts with HF in a 1:1 ratio, the moles of HF remaining after reaction are:
moles of HF remaining=0.0209mol−0.0121mol=0.0088mol
The moles of fluoride ion (F−) formed are equal to the moles of NaOH added:
moles of F−=0.0121mol
The total volume after adding NaOH is:
total volume=0.0380L+0.0220L=0.0600L
The concentrations of HF and F− are:
[HF]=0.0600L0.0088mol=0.1467M
[F−]=0.0600L0.0121mol=0.2017M
The pH can be calculated using the Henderson-Hasselbalch equation:
pH=pKa+log([HF][F−])
First, calculate pKa:
pKa=−log(7.2×10−4)=3.1427
Now, substitute the values into the equation:
pH=3.1427+log(0.14670.2017)
pH=3.1427+log(1.3751)=3.1427+0.1378=3.2805
The pH after 22.0 mL of NaOH has been added is approximately:
3.28