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NEET NEET Questions — Free MCQ Practice

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Find the $∆$ pH (initial pH –final pH) when 100 ml 0.01 M HCl is added in a solution containig 0.1 m molesof NaHCO₃ solution of negligible volume ( Ka_i =10^–7, Ka, =10^–11 for H₂CO₃) :

A.

6 + 2 log3

B.

6 – log3

C.

6 + 2Iog2

D.

6 – 21°93

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Explanation

pH of NaHCO₃ solution = 9

Now H+ + HCO₃^– → H₂CO₃

∴ no. of mmole of HCl remaining = 1 – 0.1 = 0.9 mmole

∴ pH = – log (9 × 10^–3) = – 2 log 3 + 3

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The ionization constant of benzoic acid is 6.46 x 10^–5 and Kc for silver benzoate is 2.5 x 10^–13. How many times silver benzoate is more soluble in a buffer of pH = 3.19 as compared to its solubility in pure water ?

A.

3.317

B.

9.5

C.

1000

D.

7.5

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Explanation

In pure water: C₆H₅COOAg $\overset{}{⇌}$ C₆H₅COO^– + Ag⁺

(s – x) s

C₆H₅COO^– + H₂O $\overset{}{⇌}$ C₆H₅COOH + OH^–

(s – x) x x

s (s – x) = 2.5 × 10–13 ..……….

$\frac{x^{2}}{(s - x)} = \frac{10^{- 14}}{(6 .46 \times 10^{- 5})}$ .………. (2)

Calculate s.

In buffer: C₆H₅COOAg $\overset{}{⇌}$ C₆H₅COO^– + Ag⁺

(s’ – x’) s’

C₆H₅COO^– + H⁺ $\overset{}{⇌}$ C₆H₅COOH

S’ (s’ – x’) = 2.5 × 10^–13.………. (3)

$\frac{(s' - x') \times 10^{- 3 .19}}{x'}$ = 6.46 × 10^–5.………. (4)

solve for s’.

Calculate $\frac{s'}{s} .$

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30 ml of 0.06 M solution of the protonated form of an anion acid methonine (H₂A+) is treated with 0.09 MNaOH. Calculate pH after addition of 20 ml of base. pKa 1, = 2.28 and pKa₂ = 9.2.

A.

5.5

B.

5.74

C.

9.5

D.

None

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Explanation

pH = $\frac{{pK}_{a_{1}} + {pK}_{a_{2}}}{2}$ = $\frac{2 .28 + 9 .2}{2}$ = 5.74

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A certain acid–base indicator is red in acid solution and blue in basic solution 75% of the indicator is presentin the solution in its blue form at pH = 5. Calculate the pH at which the indicator shows 90% red form?

A.

3.56

B.

5.47

C.

2.5

D.

7.4

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Explanation

pH = pK₁ + log $\frac{[{In}^{-}]}{[HIn]}$

5 = pK1 + log $\frac{75}{25}$

⇒ pK₁ = 4.523

⇒ K₁ = 3 × 10^–5

pH = 4.523 + log $\frac{10}{90}$ = 4.523 – 0.954 = 3.56

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Calculate the molar solubility of zinc tetrathiocyanato–N–mercurate (II) if its K_sp = 2.2 x 10^–7.

A.

0.00380

B.

0.000469

C.

0.0095

D.

0.0183

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Explanation

Zn [Hg(NCS)₄] $\overset{}{⇌}$ Zn⁺² + [Hg(NCS)₄]^2–

⇒ K_SP = S₂

⇒ S = $\sqrt{22 \times 10^{- 8}}$ = 4.69 × 10^–4 mol/L

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An acid–base indicator which is a weak acid has a pK_a value = 5.45. At what cocentration ratio of sodiumacetate to acetic acid would the indicator show a colour half–way between those of its acid and conjugate base forms? pK_a of acetic acid = 4.75. [log 2 = 0.3]

A.

4 : 1

B.

7 : 1

C.

5 : 1

D.

2 : 1

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Explanation

Indicator is weak acid HIN $\overset{}{⇌}$ H⁺ + In^–

given it shows colour at half way of ionization

K_a = $\frac{[H^{+}] [{In}^{-}]}{[HIn]}$ so pK_a = pK_a = pH = 5.45

but for CH₃COOh and CH₃ COONa buffer

pH = pKa + log $\frac{[Salt]}{[Acid]}$

5.45 = 4.75 + log $\frac{[Salt]}{[Acid]}$

⇒ $\frac{[Salt]}{[Acid]}$ = $\frac{5}{1}$

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The indicator constant of phenolphthalein is approximately 10^–10. A solution is prepared by adding 100.01c.c. of 0.01 N sodium hydroxide to 100.00 c.c. of 0.01N hydrochloric acid. If a few drops of phenolphthalein are now added, what fraction of the indicator is converted to its coloured form?

A.

$\frac{2}{3}$

B.

$\frac{3}{4}$

C.

$\frac{1}{2}$

D.

$\frac{9}{11}$

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Explanation

NaOH + HCl → NaCl + H₂O

mm 100.01 × 0.01 100 × 0.01

conc. $\frac{0 .001}{200}$ 0 1 1

HPh $\overset{}{⇌}$ H⁺ + Ph^–

K_a = $\frac{[H^{+}] [{Ph}^{-}]}{[HPh]}$

⇒ 10^–10 = $\frac{k_{w} [Ph]}{[{OH}^{-}] + [HPh]}$ = $\frac{10^{- 14} \times [Ph]}{\frac{0 .01}{200} [HPh]}$ $\frac{[Ph}{[HPh]} = \frac{2}{1}$

so $\frac{[{Ph}^{-}]}{[{Ph}^{-}] + [HPh]} = \frac{2}{3}$ .

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A certain mixture of HCl and CH₃ – COOH is 0.1 M in each of the acids. 20 ml of this solution is titrated against 0.1M NaOH. By how many units does the pH change from the start to the stage when the HCl is almost completely neutralised? K_a for acetic acid = 1.8 x 10^–6.

A.

2.03

B.

0.775

C.

2.325

D.

3.172

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Explanation

Initial [H⁺] = 0.1 (H⁺ from weak acid can neglect)

after neutralization of HCl concentration of CH₃COOH = $\frac{0 .1 \times 20}{40}$ = 0.05

volume would double[H⁺] = $\sqrt{K_{a} C}$

[H⁺] = $\sqrt{1 .8 \times 10^{- 5} \times 0 .05}$ = 9.48 × 10^–4

pH = 3.03

change in pH unit = 3.03 – 1 = 2.03

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A buffer solution is made by mixing a weak acid HA (K_a = 10^–6) with its salt NaA in equal amounts. What should be the amount of acid or salt that should be added to make 90 ml of buffer solution of buffer capacity. 0.1 ?

A.

10 milli moles

B.

22 milli moles

C.

9 milli moles

D.

11 milli moles

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Explanation

For buffer capacity of 0.1 we should have $(\frac{c + 0 .1}{c - 0 .1})$ = 10

where is concentration of weak acid or salt in the buffer solution

So,c + 0.1 = 10 c – 1so9c = 1.1or c = $\frac{1 .1}{9}$

So, moles required for 90 solution = $\frac{1 .1}{9}$ × 90 × 10^–3 moles = 11 milli moles.

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A sample of water has a hardness expressed as 80 ppm of Ca²⁺. This sample is passed through an ion exchange column and the Ca²⁺ is replaced by H⁺. What is the pH of the water after it has been so treated? [Atomic mass of Ca = 40]

A.

3

B.

2.7

C.

5.4

D.

2.4

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Explanation

10⁶ ml water contains 80 gm of Ca²⁺ = $\frac{80}{40}$ moles = 2 moles of Ca²⁺ = 2 × 2 moles of H⁺ ions so 10³ ml of H₂O will have = 4 × 10^–3 moles of H⁺ ions

so pH = 3 – log 4 = 3 – 0.6 = 2.4.

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