Question

Following five solution of KOH were prepare as-
First ----- 0.1 moles in 1 L
Second ----- 0.2 moles in 2 L
Third ----- 0.3 moles in 3 L
Fourth ----- 0.4 moles in 4 L
Fifth ----- 0.5 moles in 5 L
The pH of resultant solution is :-
(1) 2. (2) 1 (3) 13 (4) 7​

Answer 1

Answer:

the pH of a solution is independent term from quantity and moles

as such the pH of all the solutions will be 13

Answer 2

The pH of the resultant solution is 13

Explanation:

To calculate the molarity of the solution after mixing 5 solutions, we use the equation:

$M=\frac{n_1+n_2+n_3+n_4+n_5}{V_1+V_2+V_3+V_4+V_5}$

where,

$n_1\text{ and }V_1$ are the number of moles and volume of first KOH solution.

$n_2\text{ and }V_2$ are the number of moles and volume of second KOH solution.

$n_3\text{ and }V_3$ are the number of moles and volume of third KOH solution.

$n_4\text{ and }V_4$ are the number of moles and volume of fourth KOH solution.

$n_5\text{ and }V_5$ are the number of moles and volume of fifth KOH solution.

We are given:

$n_1=0.1mol\\V_1=1L\\n_2=0.2mol\\V_2=2L\\n_3=0.2mol\\V_3=3L\\n_4=0.4mol\\V_4=4L\\n_5=0.5mol\\V_5=5L$

Putting all the values in above equation, we get:

$M=\frac{0.1+0.2+0.3+0.4+0.5}{1+2+3+4+5}\\\\M=0.1M$

1 mole of KOH solution produces 1 mole of hydroxide ions and 1 mole of potassium ions.

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.1M$

Putting values in above equation, we get:

$pOH=-\log(0.1)\\\\pOH=1$

To calculate the pH of the solution, we use the equation:

$pH+pOH=14\\\\pH=14-1=13$

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