Question

Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.

Answer 1

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

$HCl+KOH\rightarrow KCl+H_2O$

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

$n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol$

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

$n_{KOH}=0.00768mol-0.00192mol=0.00576mol$

And the resulting concentration of KOH and OH ions as this is a strong base:

$[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M$

And the resulting pH is:

$pH=14+log(0.131)\\\\pH=13.1$

Regards!