The pH of the solution after adding 0.150 moles of solid LiF is 3.84
<u>Explanation:</u>
We have the chemical equation,
HF (aq)+NaOH(aq)->NaF(aq)+H2O
To find how many moles have been used in this
c= n/V=> n= c.V
nHF=0.250 M⋅1.5 L=0.375 moles HF
Simillarly
nF=0.250 M⋅1.5 L=0.375 moles F
nHF=0.375 moles - 0.250 moles=0.125 moles
nF=0.375 moles+0.250 moles=0.625 moles
[HF]=0.125 moles/1.5 L=0.0834 M
[F−]=0.625 moles/1.5 L=0.4167 M
To determine the problem using the Henderson - Hasselbalch equation
pH=pKa+log ([conjugate base/[weak acid])
Find the value of Ka
pKa=−log(Ka)
pH=−log(Ka) +log([F−]/[HF]
pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)
pH=-log(3.5 x 10 ^4)+log(4.996)
pH= -4.54+0.698
pH=-(-3.84)
pH=3.84
The pH of the solution after adding 0.150 moles of solid LiF is 3.84
