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Calculate the pH of a solution that is 1.0×10^{–8} M HCl.
Both water and HCl contribute significant amounts of hydronium ion, so both reactions must be considered. This problem, then, is a common–ion problem where HCl provides the common ion to the water autoionization equilibrium. So, write the two reactions, write the mass action expression for K_{w}, set up a table of concentrations, and solve. Assume 25 ^{o}C.
















Initial 



Change 



Equilibrium 


1.0×10^{–14} = [1.0×10^{–8} + x][x]
1.0×10^{–14} = 1.0×10^{–8}x + x^{2}
x^{2} + 1.0×10^{–8}x – 1.0×10^{–14} = 0
Solve with the quadratic equation to give
x = 9.5×10^{–8}, –1.1×10^{–7}
Only the positive value makes sense chemically, so
[H_{3}O^{+}] = 1.0×10^{–8} + x = 1.0×10^{–8} + 9.5×10^{–8} = 1.05×10^{–7} M
pH = –log[H_{3}O^{+}] = –log[1.05×10^{–7}] = 6.979 = 6.98 (despite the sig fig rules, pH values beyond 2 decimal places are nearly always meaningless)