Question

When 200 mL NaOH(aq) is added to 500 mL of 1.0 M HCL(aq), the pH of the resulting mixture is what?

Answer 1

Volume of NaOH = 200ml, Volume of HCl = 500ml, Strength of HCl= 1(M). Here, the strength of NaOH is not given if we consider NaOH also has same Strength as of HCl , then 200 ml NaOH neutralizes 200 ml HCl , So, excess HCl= 300ml. Now to determine the concentration of H+ ion we must follow the rule , V1 ×S1 = V2 ×S2, S = 300× 1 /(500+200) = 0.428(M) , pH = -log[H+] = -log (0.428) = 0.368. In this way we can calculate the pH. After addition of NaOH new volume is 700ml, so, the concentration in new volume is changed and can be calculated in the above way, but as you didn't supply the concentration of NaOH so I considered it as same concentration as of HCl.