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# Calculate the end point pH, when 25 mL of 0.01 mol/L HCl solution reacts exactly with 25 mL of 0.1 mol/L NH4OH solution. NH3 Kb = 1.8 x 10^-5This...

Calculate the end point pH, when 25 mL of 0.01 mol/L HCl solution reacts exactly with 25 mL of 0.1 mol/L NH4OH solution.
NH3 Kb = 1.8 x 10^-5This will be the pH of NH4Cl solution.
NH4+ + HOH ==> NH3 + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+). Solve for H3O^+, then pH = - log(H3O^+). Post your work if you get stuck. But we haven't learned how to get the pH of H3O+...we only get the pH of [H+].Same thing. They are all H3O^+ but sometimes we get a little lazy and instead of writing H3O^+ or H*H2O^+ we just omit the water part and write H^+.

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