Question

At 25 °C, how many dissociated OH– ions are there in 1243 mL of an aqueous solution whose pH is 2.07?

Answer 1

Answer: The number of $OH^-$ ions dissociated are $8.57\times 10^{11}$

Explanation:

We are given:

pH = 2.07

Calculating the value of pOH by using equation, we get:

$2.07+pOH=14\\\\pOH=14-2.07=11.93$

To calculate hydroxide ion concentration, we use the equation to calculate pOH of the solution, which is:

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

We are given:

pOH = 11.93

Putting values in above equation, we get:

$11.93=-\log[OH^-]$

$[OH^-]=10^{-11.93}=1.17\times 10^{-12}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of solution = $1.17\times 10^{-12}M$

Volume of solution = 1243 mL = 1.243 L  (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.17\times 10^{-12}M=\frac{\text{Moles of }OH^-}{1.243L}\\\\\text{Moles of }OH^-=(1.17\times 10^{-12}mol/L\times 1.243L)=1.424\times 10^{-12}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, $1.424\times 10^{-12}mol$ number of $OH^-$ will contain = $(1.424\times 10^{-12}\times 6.022\times 10^{23})=8.57\times 10^{11}$ number of ions

Hence, the number of $OH^-$ ions dissociated are $8.57\times 10^{11}$