Question

What is the pH of a solution made by mixing 200 mL of 0.0657 M NaOH, 140 mL of 0.107 M HCl, and 160 mL of H2O? I know that the answer is 2.43, but could someone please explain how? Thank you so much!!!!

Answer 1

`pH-=-log_10[H_3O^+]=-log_10(3.68xx10^-3)=-(-2.43)=2.43`

Explanation:

We assess the chemical reaction....

`HCl(aq) + NaOH(aq) rarr NaCl(s) + H_2O(l)`

We must calculate the stoichiometric equivalence of the acid and the base...

`"Moles of HCl"=140*mLxx10^-3*L*mL^-1xx0.107*mol*L^-1=0.01498*mol.`

`"Moles of NaOH"=200*mLxx10^-3*L*mL^-1xx0.0657*mol*L^-1=0.01314*mol.`

And thus, with respect to `HCl` there is a concentration of....

`((0.01498-0.01314)*mol)/((140*mL+200*mL+160*mL)xx10^-3*L*mL^-1)`

And so...`[HCl]=(1.84xx10^-3*mol)/(0.500*L)=3.68xx10^-3*mol*L^-1`...and thus `[H_3O^+]=3.68xx10^-3*mol*L^-1`

Now.............. `pH-=-log_10[H_3O^+]=-log_10(3.68xx10^-3)=-(-2.43)=2.43` .........as required...

If there is something that I have glossed over, or something that you do not follow, voice your concern, and someone will address it.... The basic definition is `pH=-log_10[H_3O^+]`; of course we first had to access `[H_3O^+]`.