Question

2055 Q. No. 10^-2

mole of KOH is dissolved in 10 litres of
water. What will be the pH of the solution?
12
Ans: pH = 11​

Answer 1

To find the pH of the solution of 0.10^-2 M KOH in 10 liters of water, calculate the molarity, then use the water dissociation constant to determine [H3O+], and finally, take the negative log to find the pH, which is 12.

Step-by-step explanation:

To determine the pH of a solution when 0.10-2 mole of KOH is dissolved in 10 litres of water, we first need to calculate the molarity of KOH. Since 0.10-2 moles are dissolved in 10 liters of water, the molarity ([OH-]) is 0.10-2 M.

We use the water dissociation constant (Kw) which at 25 °C is 10-14. The concentration of hydronium ions [H3O+] can be calculated by dividing Kw by the concentration of OH- ions, giving [H3O+] = 10-14 / 0.10-2 M = 10-12 M. Then, to find the pH, we take the negative logarithm (base 10) of the [H3O+], which gives us pH = -log(10-12) = 12.

Therefore, the pH of the solution is 12.

Answer 2

11

Step-by-step explanation:

Moles of KOH =
`10^(-2)`

Volume of water = 10 liters

Concentration of KOH is given by

`[KOH]=(10^(-2))/(10)\\\Rightarrow [KOH]=10^(-3)\ \text{M}`

`[KOH]` is strong base so we have the following relation

`[KOH]=[OH^(-)]=10^(-3)\ \text{M}`

`pOH=-\log [OH^(-)]=-\log10^(-3)`

`\Rightarrow pH=14-3=11`

So, pH of the solution is 11