Question 11 Not yet answered Marked out of 1.00 Calculate how many mL of 0.100 M

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Question 11 Not yet answered Marked out of 1.00 Calculate how many mL of 0.100 M NaOH are needed to neutralize completely 93.0 mL of 0.0600 M acetic acid Answer Flag question Question 12 Not yet answered Marked out of 1.00 Titration of Strong Acid Calculate the pH during the titration of 100.0 mL of 0.200 M HCI with 0.400 M NaOH. First what is the initial pH (before any NaOH is added)? Answer: Flag question Question 13 What is the pH after 34.0 mL of NaOH are added? Not yet answered Marked out of 1.00 Answer: Flag question Question 14 What is the pH after 50 mL of NaOH are added? Not yet answered Marked out of 1.00 Answer: Flag question Question 15 Not yet answered Marked out of 1.00 What is the pH after 65.3 mL of NaOH are added? Answer:

Explanation / Answer

11)
Balanced chemical equation is:
CH3COOH + NaOH ---> CH3COONa + H2O


Here:
M(CH3COOH)=0.06 M
M(NaOH)=0.1 M
V(CH3COOH)=93.0 mL

According to balanced reaction:
1*number of mol of CH3COOH =1*number of mol of NaOH
1*M(CH3COOH)*V(CH3COOH) =1*M(NaOH)*V(NaOH)
1*0.06 M *93.0 mL = 1*0.1M *V(NaOH)
V(NaOH) = 55.8 mL
Answer: 55.8 mL

12)
[H+] = 0.200 M

use:
pH = -log [H+]
= -log (0.2)
= 0.699

Answer: 0.699

13)
Given:
M(HCl) = 0.2 M
V(HCl) = 100 mL
M(NaOH) = 0.4 M
V(NaOH) = 34 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.2 M * 100 mL = 20 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.4 M * 34 mL = 13.6 mmol


We have:
mol(HCl) = 20 mmol
mol(NaOH) = 13.6 mmol
13.6 mmol of both will react
remaining mol of HCl = 6.4 mmol
Total volume = 134.0 mL

[H+]= mol of acid remaining / volume
[H+] = 6.4 mmol/134.0 mL
= 0.0478 M


use:
pH = -log [H+]
= -log (4.776*10^-2)
= 1.321

14)
Given:
M(HCl) = 0.2 M
V(HCl) = 100 mL
M(NaOH) = 0.4 M
V(NaOH) = 50 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.2 M * 100 mL = 20 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.4 M * 50 mL = 20 mmol
Since mol of acid and base are equal. pH will be 7
pH = 7

15)
Given:
M(HCl) = 0.2 M
V(HCl) = 100 mL
M(NaOH) = 0.4 M
V(NaOH) = 65.3 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.2 M * 100 mL = 20 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.4 M * 65.3 mL = 26.12 mmol


We have:
mol(HCl) = 20 mmol
mol(NaOH) = 26.12 mmol
20 mmol of both will react

remaining mol of NaOH = 6.12 mmol
Total volume = 165.3 mL

[OH-]= mol of base remaining / volume
[OH-] = 6.12 mmol/165.3 mL
= 0.037 M


use:
pOH = -log [OH-]
= -log (3.702*10^-2)
= 1.4315


use:
PH = 14 - pOH
= 14 - 1.4315
= 12.6

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