NTA Abhyas 2020
Exam NTA Abhyas
Year 2020
Subject Chemistry
Chapter Ionic Equilibrium NTA Abhyas 2020 · Chemistry · Ionic Equilibrium
One litre of 1M solution of an acid HA ( K _ a = 10 ^ - 4 at 25 ^ o...
One litre of
1M \text{1}\text{M} 1 M solution of an acid
HA ( K a = 10 − 4 at 25 o C ) h a s pH=2. I t i s d i l u t e d b y w a t e r s o t h e n e w \text{HA} \left({K}_{a} \text{=}{\text{10}}^{- \text{4}} \text{at 25}{}^{o} \text{C}\right) has \text{pH}\text{=}\text{2}\text{.} It is diluted by water so the new HA ( K a = 10 − 4 at 25 o C ) ha s pH = 2 . I t i s d i l u t e d b y w a t erso t h e n e w pH
b e c o m e s d o u b l e . T h e s o l u t i o n w a s d i l u t e d t o becomes double. The solution was diluted to b eco m es d o u b l e . T h eso l u t i o n w a s d i l u t e d t o y
× \times × 1{0}^{z}ml
. T h e v a l u e o f y + z 2 . The value of \dfrac{y + z}{2} . T h e v a l u eo f 2 y + z is:
Step-by-step solution
p H = 2 ⇒ [ H + ] 1 = 10 − 2 = c 1 α 1 = 1 × α 1 pH = 2 ⇒ {\left[{H}^{+}\right]}_{1} = {10}^{- 2} = {c}_{1}{\alpha}_{1} = 1 \times {\alpha}_{1} p H = 2 ⇒ [ H + ] 1 = 10 − 2 = c 1 α 1 = 1 × α 1 α 1 = 10 − 2 {\alpha}_{1} = {10}^{- 2} α 1 = 10 − 2 p H = 4 ⇒ [ H + ] 2 = 10 − 4 = 1 × 1000 V × α 2 pH = 4 ⇒ {\left[{H}^{+}\right]}_{2} = {10}^{- 4} = \dfrac{1 \times 1000}{V} \times {\alpha}_{2} p H = 4 ⇒ [ H + ] 2 = 10 − 4 = V 1 × 1000 × α 2 k a = 10 − 4 = α 2 2 C 2 1 − α 2 = α 2 × α 2 × C 2 1 − α 2 {k}_{a} = {10}^{- 4} = \dfrac{{\alpha}_{2}^{2}{C}_{2}}{1 - {\alpha}_{2}} = \dfrac{{\alpha}_{2} \times {\alpha}_{2} \times {C}_{2}}{1 - {\alpha}_{2}} k a = 10 − 4 = 1 − α 2 α 2 2 C 2 = 1 − α 2 α 2 × α 2 × C 2 10 − 4 = α 2 × 10 − 4 1 − α 2 ⇒ α 2 = 0.5 {10}^{- 4} = \dfrac{{\alpha}_{2} \times {10}^{- 4}}{1 - {\alpha}_{2}} ⇒ {\alpha}_{2} = 0 .5 10 − 4 = 1 − α 2 α 2 × 10 − 4 ⇒ α 2 = 0.5 [ H + ] 2 = 1 × 1000 V × α 2 = 10 − 4 {\left[{H}^{+}\right]}_{2} = \dfrac{1 \times 1000}{V} \times {\alpha}_{2} = {10}^{- 4} [ H + ] 2 = V 1 × 1000 × α 2 = 10 − 4 1 × 1000 V × 0.5 = 10 − 4 \dfrac{1 \times 1000}{V} \times 0 .5 = {10}^{- 4} V 1 × 1000 × 0.5 = 10 − 4 V = 5 × 10 6 m l V = 5 \times {10}^{6}ml V = 5 × 10 6 m l ⇒ x + y 2 = 5 + 6 2 = 11 2 = 5.5 ⇒ \dfrac{x + y}{2} = \dfrac{5 + 6}{2} = \dfrac{11}{2} = 5 .5 ⇒ 2 x + y = 2 5 + 6 = 2 11 = 5.5
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