# What will be the pH of the solution that results from the addition of 20mL of 0.01M Ca(OH)2 to 30mL of 0.01M HCl?

Apr 5, 2017

$p H = 2.69897000434$

#### Explanation:

It is a strong acid-base reaction as HCl is a strong acid and Ca(OH)2 is a strong base

$C a {\left(O H\right)}_{2} + H C l = C a C {l}_{2} + 2 {H}_{2} O$

The net ionic reaction is

${H}^{+} \left(a q\right) + O {H}^{-} \left(a q\right) = {H}_{2} O \left(l\right)$

The number ${H}^{+}$ ions in the solution is the concentration of HCl because strong acids dissociate completely thus producing
$\text{1mole of H+"/"1mole of acid.}$ This is only applicable to strong acids. The same is the case with strong base but instead of ${H}^{+}$ they produce strong conjugate bases like $O {H}^{-}$

HCl is a strong acid thus

moles of HCl = moles of ${H}^{+}$

Calculate no. of moles in 30mL of 0.01M HCl

$L \cdot M = \text{moles}$

$30 \cdot {10}^{-} 3 L \cdot 0.01 = \text{0.0003moles}$

0.0003moles of HCl will produce 0.0003moles of ${H}^{+}$

Now calculate the number of moles in 20mL of 0.01M $C a {\left(O H\right)}_{2}$ solution.

$C a {\left(O H\right)}_{2} i s a s t r o n g b a s e t h u s$

moles of $C a {\left(O H\right)}_{2}$ = moles of $O {H}^{-}$

$20 \cdot {10}^{-} 3 L \cdot 0.01 M = \text{0.0002moles}$

0.0002moles of $C a {\left(O H\right)}_{2}$ = 0.0002moles of $O {H}^{-}$

As this is titrating of a strong acid with a strong base the addition of the base results in decrease in ${H}^{+}$ ions. I mol of OH- results in decrease of 1 mol of H+

0.0003mole - 0.0002mole = 0.0001mole of ${H}^{+}$

You must now calculate the pH by using the moles and the total volume

But while calculating pH we must use the molarity and not mole.
For that we need to add both the volumes that is

20mL + 30mL = 50mL = 0.05L

$\text{0.0001mole"/"0.05L} = \frac{x}{1 L}$

= $\frac{0.002}{1 L}$

= 0.002M of ${H}^{+}$

$p H = - \log \left({H}^{+}\right)$

$p H = 2.69897000434$