# 204 - Chemistry Questions Answers

100 c.c. of N/10 NaOH solution is mixed with 100 c.c. of N/5 HCl solution and the whole volume is made to 1 litre. The pH of the resulting solution will be

Joshi sir comment

mili gm eq of NaOH = NV = 10

mili gm eq of HCl = 20

total mili gm eq = 20-10 = 10 of HCl

so N = 10/200 = 0.05

so pH = log[1/0.05]

When 0.1 mole of ammonia is dissolved in sufficient water to make 1 litre of solution. The solution is found to have a hydroxide ion concentration of 1.34 x 10-3 . The dissociation constant of ammonia is

Joshi sir comment

NH3 + H2O---------------------> NH4+ + OH-

0.1                                      0           0

0.1(approx)                      1.34*10-3  1.34*10-3

now calculate dissociation constant

Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl. Ka of acetic acid is 1.8 x 10-5

Joshi sir comment

CH3COOH ----------------> CH3COO- + H+

0.01                                        0                0.1

0.01-x                                      x               0.1+x

0.1+x =0.1 for x is negligibly small

so K= x*0.1/(0.01-x)

now solve

The dissociation constant of a weak acid  HA and weak base BOH are 2 x 10-5 and 5 x 10-6 respectively. The equilibrium constant for the neutralization reaction of the two is (ignore hydrolysis of resuting salt)

Joshi sir comment

HA + BOH ------------------>  B+ + A- + H2O

K = [B+][A-][H2O]/[HA][BOH]

= [B+][OH-]/[BOH] *  [H+][A-]/[HA] *  [H2O]/[H+][OH-] = Kb *  Ka  /  Kw

now solve

Which of the following increasing order of pH of 0.1 M solution of the component    (a) HCOONH4  (b) CH3COOHNH4  (c) CH3COONa (d) NH4Cl  is correct

Joshi sir comment

(d) NH4Cl   <     (a) HCOONH4    <    (b) CH3COONH4     <      (c) CH3COONa

Which of the following molar ratio of NH3 and HCl in aqueous solution will constitute a buffer ? (a) 1 : 2 (c) 1 : 1 (b) 1 : 3 (d) 2 : 1

Joshi sir comment

in (d) after reaction no. of moles of NHwill be 1 and 1 mole NH4Cl will be formed so it will be a basic buffer.

In order to prepare a buffer of pH 8.26, the amount of (NH4 )2SO4 required to be mixed with 1L of 0.1M NH3 (pKb of NH= 4.74) is

Solution by Joshi sir

NH3 + H2 ------------>  NH4+ + OH-

so Kb = [NH4+] [OH-]/[NH3][H2O]

so Kb = [NH4+]/[NH3][H+] * [H+][OH-]/[H2O]

so Kb = [NH4+]/[NH3][H+] * Kw

now take log and solve. concentration of  (NH4 )2SO4 will be double of [NH4+]

if you find any problem on solving this then reply

1.0 ml of dilute solution of NaOH is added to 100 ml of a buffer of pH 4. The pH of the resulting solution

(a) becomes 7.0

(b) becomes 9.0

(c) becomes 3.0

(d) Remains practically unchanged

Joshi sir comment

it is strong acidic buffer so answer is (d)

10 ml of M/200 H2SO4 is mixed with 40 ml of M/200 H2SO4. The ph of the resulting solution is

Joshi sir comment

M of H2SO4 = 1/200

so [H+] = 2*1/200 = 1/100 = 10-2

now calculate pH

The pH of solution at 25°C which has twice as many hydroxide ion as in pure water at 25°C , will be

Joshi sir comment

no of hydroxide ion in pure water = 10-7

so according to the given condition no. of hydroxide ions in solution = 2*10-7

so pOH = -log2*10-7

so pH = 14 - pOH