1.The strong acid, HCl, is titrated with a strong base, NaOH, in aqueous solutio
ID: 1033878 • Letter: 1
Question
1.The strong acid, HCl, is titrated with a strong base, NaOH, in aqueous solution. What is the pH of the titration solution after the addition of 15.31 mL of 0.100 M NaOH to a 50.00 mL of 0.100 M HCl ?
2. A weak acid, HA, is titrated with a strong base, NaOH. If 25.00 mL of the 0.1000 M HA is titrated with 19.95 mL of 0.1000 M NaOH, what is the pH of the titration solution?
For HA, Ka = 1.8 x 10-6
3. Monoprotic weak acid, HA, is titrated with strong base, NaOH, in aqueous solution. If 25.00 mL of 0.100 M HA is titrated with 0.100 M NaOH, what is the pH of the titration solution after the addition of 29.15 mL of NaOH?
Explanation / Answer
1)
Given:
M(HCl) = 0.1 M
V(HCl) = 50 mL
M(NaOH) = 0.1 M
V(NaOH) = 15.31 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.1 M * 50 mL = 5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 15.31 mL = 1.531 mmol
We have:
mol(HCl) = 5 mmol
mol(NaOH) = 1.531 mmol
1.531 mmol of both will react
remaining mol of HCl = 3.469 mmol
Total volume = 65.31 mL
[H+]= mol of acid remaining / volume
[H+] = 3.469 mmol/65.31 mL
= 0.0531 M
use:
pH = -log [H+]
= -log (5.312*10^-2)
= 1.2748
Answer: 1.27
2)
Given:
M(HA) = 0.1 M
V(HA) = 25 mL
M(NaOH) = 0.1 M
V(NaOH) = 19.95 mL
mol(HA) = M(HA) * V(HA)
mol(HA) = 0.1 M * 25 mL = 2.5 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 19.95 mL = 1.995 mmol
We have:
mol(HA) = 2.5 mmol
mol(NaOH) = 1.995 mmol
1.995 mmol of both will react
excess HA remaining = 0.505 mmol
Volume of Solution = 25 + 19.95 = 44.95 mL
[HA] = 0.505 mmol/44.95 mL = 0.0112M
[A-] = 1.995/44.95 = 0.0444M
They form acidic buffer
acid is HA
conjugate base is A-
Ka = 1.8*10^-6
pKa = - log (Ka)
= - log(1.8*10^-6)
= 5.745
use:
pH = pKa + log {[conjugate base]/[acid]}
= 5.745+ log {4.438*10^-2/1.123*10^-2}
= 6.341
Answer: 6.34
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