Question

How many moles of H3O+ or OH- must you add to a liter of strong base solution to adjust its pH from 9.350 to 7.760? Assume a negligible volume change

Answer 1

The given initial pH = 9.35 , this represents there are more number of OH- ions in aqueous solution. The final pH = 7.76 also represents still the solution is basic.

Let us calculate first OH- moles

pH + pOH = 14

pOH = 14 - pH = 14 - 9.35 = 4.65

[OH-] = 10^-pOH = 10^-4.65 = 0.00002238721 moles

target pH = 7.76, same way above gives

pOH = 14 - pH = 14 - 7.76 = 6.24

[OH-] = 10^-pOH = 10^-6.24 = 5.75 x 10^-7 moles

[OH-] = 5.75 x 10^-7 moles

one mol H+ neutralizes one mol OH- ion and forms H₂O thereby reduces pH from 9.35 to 7.76

H⁺ + OH^- --> H₂O

moles OH- has to be neutralized = 0.00002238721 moles - 5.75 x 10^-7 moles = 0.00002181177 moles

H₃O⁺ has to be added = 0.00002181 moles

moles of H₃O⁺ has to be added = 0.00002181 moles ( 2.181 x 10^-5 mol)

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