Question

0.001 mole of strong electrolytes zn(oh)2 is present in 200 ml of an aqueous solution. The ph of this solution is.

Answer 1

is the correct answer 3.25

Answer 2

pH of the solution is 12

Explanation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$  

Molarity of $Zn(OH)_2$ solution = ?

moles of  $Zn(OH)_2$  = 0.001

Volume of solution = 200 mL

Putting values in equation 1, we get:

$\text{Molarity of the solution}=\frac{0.001\times 1000}{200ml}=0.005M$

1 M $Zn(OH)_2$ gives = 2M  $OH^-$

Thus $0.005$ M $Zn(OH)_2$ gives= $\frac{2}{1}\times 0.005=0.01$M of $OH^-$

$pOH=-log[OH^-]$

$pOH=-log[0.01]=2$

$pH+pOH=14$

$pH=14-2=12$

pH of the solution is 12.

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