Question

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

Answer 1

pH = 13.1

Step-by-step 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^-]=(0.00576mol)/(0.012L+0.032L)=0.131M`

And the resulting pH is:

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

Regards!