A solution is prepared by mixing 88.0 mL of 5.00 M HCl and 26.0 mL of 8.00 M HNO3. Water is then added until the final volume is 1.00 L. How would you calculate [H+], [OH -], and the pH for this solution? please explain?

1 Answer
Nov 12, 2015

Answer:

This is a typical acid/base equilibrium problem, that involves the use of logarithms.

Explanation:

We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric #H_3O^+#.

Moles of nitric acid: #26.0xx10^(-3)*L xx 8.00*mol*L^-1 = 0.208*mol # #HNO_3(aq)#.

And, moles of hydrochloric acid: #88.0xx10^(-3)*L xx 5.00*mol*L^-1 = 0.440*mol# #HCl(aq)#.

This molar quantity is diluted to #1.00# #L#. Concentration in moles/Litre = #((0.208+0.440)* mol)/(1L)# #=# #0.648*mol*L^(-1).#

Now we know that water undergoes autoprotolysis:
#H_2O(l) rightleftharpoons H^+ +OH^-#. This is another equilibrium reaction, and the ion product #[H^+][OH^-]# #=# #K_w#. This constant, #K_w# #=# #10^(-14)# at #298K#.

So #[H^+] =0.648*mol*L^(-1)#; #[OH^-] = K_w/[[H^+]]# #=# #(10^-14)/0.648# #=# #??#

#pH# #=# #-log_(10)[H^+] = -log_(10)(0.648) = ??#

Alternatively, we know further that #pH + pOH =14#. Once you have #pH#, #pOH# is easy to find. Take the antilogarithm of this to get #[OH^-]#.