1) Give the pH for the following solution below... Give the pH for the following
ID: 955905 • Letter: 1
Question
1) Give the pH for the following solution below...
Give the pH for the following solutions below A buffer is prepared by making a 750.00 mL of a solution that is both 0.300 M solution of hypochloric acid and 0.250M sodium hypochorite, what is the pH of this solution after 5.0 mL of 5.0 M HCI has been added given Ka for hypochloric acid = 3.0 Times 10^-8 The titration of 450.00 mL of 0.5 M nitric acid with 0.3 M sodium hydroxide after 250.0 mL of sodium hydroxide has been added Given the K_b for aniline = 4.3 Times 10^-10 at 25 degree C. What is the pH of a 0.40 Molar solution of anilinium nitratate?Explanation / Answer
(A)
0.3 M HOCl contains 0.3 moles of HOCl
0.25 M NaOCl contains 0.25 moles of HOCl
These are in 750 mL (= 0.75 L)
[HOCl] = 0.3 mole / 0.75 L = 0.4 M
[NaOCl] = 0.25 mole / 0.75 L = 0.33 M
Ka HOCl = 3.0 x 10-8
pKa = - log(ka) = -log (3.0 x 10-8) = 7.52
Now, using the Henderson - Hasselbalch equation:
pH =pKa + log {[salt] / [acid]
= 7.52 + log (0.33/0.40)
= 7.52 + log (0.825)
= 7.52 -0.084
= 7.436
The added HCl will react with the NaOCl in solution , and increase the HOCl NaOCl concentration and reduce the NaOCl concentration
5 mL of 5.O M HCl = (5 mole/L) x (0.005) L = 0.025 mole
HCl reacts with NaOCl in 1:1 molar ratio.
Mole of HOCl = 0.3 mole + 0.025 mole = 0.325 mole
Mole of NaOCl = 0.25 mole - 0.025 mole = 0.225 mole
[HOCl] = (0.325 mole/ 0.75 L) = 0.43 M
[NaOCl] = (0.225 mole / 0.75 L) = 0.30 M
Now, using the Henderson - Hasselbalch equation:
pH =pKa + log {[salt] / [acid]
= 7.52 + log (0.30/0.43)
= 7.52 + log (0.70)
= 7.52 -0.155
= 7.365
(B)
Titration of a strong acid with a strong base:
Volume of 0.50 M HNO3 = 450 mL
So, moles of HNO3 = 0.5 M x 0.45 L = 0.225 moles
volume of 0.30 M NaOH = 250 mL
So, moles of NaOH = 0.3 M x 0.25 L = 0.075 moles
Since number of moles of HNO3 is more, so excess of moles of HNO3 = (0.225 – 0.075) moles = 0.15 moles.
Now the total volume = 450 mL + 250 mL = 700 mL = 0.7 L
Hence, [HNO3] = (0.15 mole) / (0.7 L) = 0.214 M
[H+] = 0.21 M
pH =-log[H+] = -log (0.21) = 0.7
(C)
Anilinium nitrate is a salt of weak base and strong acid ....
so its pH will be given by :
pH = 1/2 [ pKw -pKb -log C ]
pKw = -log Kw where Kw is dissociation constant of water
Kw = 10-14 so
pKw = 14
Given C = 0.4 M....
so log C = -0.4
Kb = 4.3 x 10-10.....
so pKb = -log (4.3 x 10-10) = 9.36
putting in pH expression : pH = 1/2 [ 14 - 9.36 - (-0.4)] = 1/2 [5.04] = 2.52
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