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Find out pH and pOH of a 0.0235 M HCl solution at Everett Community College Tutoring Center Student Support Services Program.
Typology: Exercises
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Everett Community College Tutoring Center Student Support Services Program
Solutions Note: The significant figures in the concentration of [H+] or [OH - ] is equal to the number of decimal places in the pH or pOH and vice versa.
pH = -log[H+] = -log(0.0235) = 1.
pH = -log[H+] = -log(0.0235) = 1. pOH = 14.000 – pH = 14.000 – 1.629 = 12.
pOH = -log[OH-] = -log(6.50 x 10-3) = 2. pH = 14.000 – pOH = 14.000 – 2.187 = 11.
Since there is both acid and base we will assume a 1 mole acid:1 mole base ratio of neutralization. There is more base than acid so the leftover base is what will affect the pH of the solution. 3.60 x 10-3^ moles - 5.95 x 10-4^ moles = 3.01 x 10-3^ moles NaOH 3.01 x 10-3^ moles NaOH = 3.01 x 10-3^ M NaOH 1.00 L soln pOH = -log[OH-] = -log(3.01 x 10-3) = 2. pH = 14.000 – pOH = 14.000 – 2.521 = 11.
pOH = -log[OH-] = -log(6.2 x 10-5) = 4.
pH = 14.00 – pOH = 14.00 – 4.21 = 9.
pH = -log[H+] = -log(1.00 x 10-7) = 7. pOH = 14.000 – pH = 14.000 – 7.000 = 7. pOH = -log[OH-] = -log(OH-) = 7.000 we can use this to find the OH-^ concentration -log[OH-] = 7. log[OH-]-1^ = 7. log[OH-]-1^ 7. 10 = 10 [OH-]-1^ = 107. 1 = 107. [OH-]
[OH-] = 1.00 x 10-7^ M The concentrations of H+^ and OH-^ are equal, as are the pH and pOH, so the solution must be neutral.