One litre of 1 M solution of an acid HA(Ka = (10)- 4 at 25 ° C) has pH=2. It is diluted by water so the new pH becomes double. The solution was diluted to y× 10z ml . The value of y+z is:

Question

One litre of 1 M solution of an acid HA(Ka = (10)- 4 at 25 ° C) has pH=2. It is diluted by water so the new pH becomes double. The solution was diluted to y× 10z ml . The value of y+z is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Ostwald's dilution law describes the dissociation constant of the weak electrolyte with the degree of dissociation

(α)

and the concentration of the weak electrolyte.
Dissociation of weak acid:
Solution

p H = 2 \Rightarrow [H^{+}]_{1} = 1 0^{- 2} = c_{1} α_{1} = 1 \times α_{1}

α_{1} = 1 0^{- 2}

p H = 4 \Rightarrow [H^{+}]_{2} = 1 0^{- 4} = \frac{1 \times 1000}{V} \times α_{2}

k_{a} = 1 0^{- 4} = \frac{α _{2}^{2} C _{2}}{1 - α _{2}} = \frac{α _{2} \times α _{2} \times C _{2}}{1 - α _{2}}

1 0^{- 4} = \frac{α _{2} \times 1 0 ^{- 4}}{1 - α _{2}} \Rightarrow α_{2} = 0.5

[H^{+}]_{2} = \frac{1 \times 1000}{V} \times α_{2} = 1 0^{- 4}

\frac{1 \times 1000}{V} \times 0.5 = 1 0^{- 4}

V = 5 \times 1 0^{6} m l

y + z = 5 + 6 = 11

Q. One litre of 1M solution of an acid HA(Ka​=(10)−4 at 25∘C) has pH=2. It is diluted by water so the new pH becomes double. The solution was diluted to y×10zml . The value of y+z is:

Q. One litre of $1 M$ solution of an acid $H A (K_{a} = (10)^{- 4}$ at $2 5^{\circ} C)$ has $p H = 2.$ It is diluted by water so the new $p H$ becomes double. The solution was diluted to $y \times 1 0^{z} m l$ . The value of $y + z$ is: