Calculate the pH of each of the following strong acid solutions.
A. 6.80 mL of 1.20 M HCl diluted to 0.400 L
B. A mixture formed by adding 41.0 mL of 2.5×10−2 M HCl to 160 mL of 1.5×10−2 M HI
(a) Dilution formula,
M1V1 = M2V2
1.20 * 6.80 = M2 * 400
M2 = Final concentration of HCl solution =0.0204 M
HCl (aq.) ---------------> H+ (aq.) + Cl- (aq.)
[H+] = [HCl] = 0.0204 M
pH = - Log[H+] = - Log(0.0204) = 1.69
(B)
Both HCl and HI are strong acids hence they undergo complete ionization.
Molarity of resulting solution, M = (M1V1 + M2V2) / (V1+V2)
M = (2.5*10-2*41.0 + 1.5*10-2*160) / (41.0 + 160)
M = 0.0170 M
Since both HI and HCl are mono protic acids,
[H+] = 0.0170 M
pH = - Log[H+] = - Log(0.0170) = 1.77
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