Consider the titration of 43.6 mL of 0.230 M HF with 0.190 M NaOH. Calculate the
ID: 894083 • Letter: C
Question
Consider the titration of 43.6 mL of 0.230 M HF with 0.190 M NaOH. Calculate the pH at each of the following points. Could you please show step by step and the answer.
A)How mL of base re required to reach the equivalence point? Answer 52.8 mL
B) Calculate the pH after the addition of 10.9 mL of base. My asnwers were pH 5.85, 1.99, 2.00 and 2.6 but were incorrect
C) Calculate the pH at the halfway to the equivalence point.
D)Calculate the pH at the equivalence point
E)Calculate the pH after the additon of 10.9 mL of base.
Explanation / Answer
This is titration between strong acid and strong base
A)
Use neutralisation formula
M1*V1 = M2*V2
1 is HF and 2 is NaOH
0.23*43.6 = 0.19*V
V=52.8 mL
Answer:52.8 mL
B)
Number of moles of HF = M*V = 0.23 M * 0.0436 L =0.010028 mol
Number of moles of NaOH = M*V = 0.19 M * 0.0109 L =0.002071 mol
0.002071 mol of each will react.
Remaining HF = 0.010028 - 0.002071 =0.007957 mol
Total volume = 0.0436 + 0.0109 L = 0.0545 L
[HF] = number of moles/ Volume
= 0.007957 / 0.0545
=0.146 M
pH = -log [H+]
=-log (0.146)
= 0.84
C)
At halfway equivalence point volume of base added =52.8/2 =26.4 mL
Number of moles of HF = M*V = 0.23 M * 0.0436 L =0.010028 mol
Number of moles of NaOH = M*V = 0.19 M * 0.0264 L =0.005016 mol
0.005016 mol of each will react.
Remaining HF = 0.010028 - 0.005016 =0.005012 mol
Total volume = 0.0436 + 0.0264L = 0.07 L
[HF] = number of moles/ Volume
= 0.005012 / 0.07
=0.0716 M
pH = -log [H+]
=-log (0.0716)
= 1.15
D)
At equivalence point equal moles of each will react.
Hence ph =7
E)
Already done in part B