Question

PROBLEMS 1) Acids and bases: Calculate the following a) [H30] in a 0.500 M solution of HCl b) pH of a 0.100 M solution of HCI c) pH of a 0.100 M solution of HNO3 d) [OH] in a 0.500 M solution of HClnote, for this you need to use Kw e) pH of a 0.100 solution of NaOH (also need to use some form of Kw here)

Answer 1

a) [H_3 O^+] in a 0.500 M solution of HCl. 0.500 M of HCl. [H_3 O^+] rightwardsdoublearrow [H^+] = 0.500 M b) p^H of a 0.100 M solution of HCl [H^+] = 0.100 M p^H = -log [H^+] = -log [0.100] p^H = -(1) p^H = 1 c) p^H of a 0.100 M solution of HNO_3. [H^+] = 0.100 M p^H = -log [H^+] = -(-1) p^H = 1 d) [OH^-] in 0.500 M solution of HCl [H^+] = 0.500 M p^H = -log (0.500) = 0.30 Therefore K_w = [H^+] [OH^-] = 1.0 times 10^-14 mol^2 ^-6. Therefore p^H + p^OH = p^K_w p^H + p^OH = 14 Therefore p^OH = 14 - p^H p^OH = 14 - 0.30 p^OH = 13.7 \r\n Therefore p^OH = -log (OH) 13.7 = -log [OH] Therefore [OH^-] = 1/10^13.7 [H^-] = 1.99 times 10^-14 e) p^H of a 0.100 M solution of NaOH [OH^-] = 0.100 M Therefore p^OH = -log (0.100 M) = -(-1) p^OH = 1