QUESTION 3 35.0 mL of a weak acid was titrated to the equivalence point with 28.

Question

QUESTION 3

35.0 mL of a weak acid was titrated to the equivalence point with 28.0 mL of 0.100 M NaOH. The pH was monitored during the titration. At 14.0 mL of base added, the pH was found to be 2.73. What is the Ka of the weak acid?
(NOTE: To enter answer in scientific notation, use E instead of x10. For example, 1.0x10-3 should be entered as 1.0E-3.)

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QUESTION 4

The weak base hydroxylamine, HONH2, is titrated with HCl. What is the pH after 9.05 mL of 0.20 M HCl is added to 30.0 mL of 0.20 M HONH2? Kb of HONH2 = 1.1 x 10-8 (Give your answer to two decimal places.)

Using the titration curve shown below, at what volume of acid added will pH = pKa? Using the titration curve shown below, at what volume of acid added will pH = pKa? 25 mL of 0.20 M unknown acid is titrated with 0.20 M NaOH, giving the following titration curve. Which of these statements about this titration is true?

Explanation / Answer

5)
pH = pKA at steep point which correspond to pH=4
at pH= 4 volume = 100 mL
Answer: 100 mL

3)
Let weak acid be HA
At equivalence point, number of moles of acid equal number of moles of base
So, total number of moles of acid present = 0.1M * 0.028 L=2.8*10^-3 mol

When 14 mL of 0.1 M base is added
Number of moles of base added= 0.1*0.014= 1.4*10^-3 mol
1.4*10^-3 mol of acid and base will react as :
HA + OH- ---> A- + H2O
moles of A- formed = 1.4*10^-3 mol
moles of HA remaining = 2.8*10^-3 mol -1.4*10^-3 mol = 1.4*10^-3 mol
Total volume = 35 mL+ 14 mL = 49 mL = 0.049 L
[A-] = numbe of moles / volume = (1.4*10^-3)/0.049 =0.029 M
[HA] = numbe of moles / volume = (1.4*10^-3)/0.049 =0.029 M

use:
pH = pKa+ log {[A-]/[HA]}
2.73 = pKa + log (0.029/0.029)
2.73 = pKa + 0
pKa =2.73
use:
pKa = -log Ka
2.73 = -log Ka
Ka= 1.86*10^-3
Answer: 1.86*10^-3

I am allowed to answer only 1 question at a time. But I have answered 2 since 1st one was very easy

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