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

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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

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Joshi sir comment

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

0.1                                      0           0

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

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

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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)

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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

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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 

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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

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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

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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

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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

Asked By: SWATI KAPOOR
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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 

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