Calculate the pH of a solution obtained by mixing 477 mL of 0.17 M hydrochloric acid with 253 mL of M lithium hydroxide. Assume the combined volume is the sum of the two original volumes.

pH = 0.984

Explanation:

Molarity LiOH = 2.1x10⁻²M

HCl will react with LiOH as follows:

HCl + LiOH → H₂O + LiCl

1 moles of HCl reacts per mole of LiOH

Moles of each reactant in solution are:

HCl = 0.477L ₓ (0.17mol / L) = 0.08109 moles HCl

LiOH = 0.253L ₓ (2.1x10⁻² mol / L) = 5.313x10⁻³ moles of LiOH.

That means LiOH is the limiting reactant and excess moles of HCl that will remain in solution are:

0.08109 mol - 0.005313mol = 0.0758 moles HCl

As HCl dissociates in water as H⁺ and Cl⁻ ions, you will have in solution 0.0758 moles of H⁺

pH = -log [H⁺] and [H⁺] = moles H⁺ / L of solution.

Volume of the mixture in liters is: 0.477L + 0.253L = 0.730L.

That means [H⁺] is 0.0758 moles of H⁺ / 0.730L = 0.1038M

Replacing:

pH = -log [H⁺]

pH = -log [0.1038]

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