What is the molarity of a KOH(aq) solution, if a 50.0 mL sample is fully neutralized by 28.7 mL of 0.3049 M HI(aq)?

How many mL of 0.45M HCl(aq) are needed to neutralize a solution of 144.00mL of 0.35M barium hydroxide?

1. The reaction equation: KOH (aq) + HI (aq) => KI (aq) + H2O (l)

The amount of HI neutralized is 0.3049 mol/L * 0.0287 L = 8.751 * 10^{-3} mol

The coefficients at KOH and HI are equal, so the amount of KOH neutralized is also 8.751 * 10^{-3} mol.

The molarity of its solution is 8.751 * 10^{-3} mol / 0.0500 L = 0.175 M.

2. The reaction equation is Ba(OH)2 (aq) + 2HCl (aq) => BaCl2 (aq) + 2H2O (l)

The amount of Ba(OH)2 neutralized is 0.35 mol/L * 0.1440 L = 0.0504 mol

The amount of HCl is twice as much (ratio of the coefficients is 2:1) and equals 2 * 0.504 = 0.1008 mol

The volume of the acid solution is 0.1008 mol / 0.45 mol/L = 0.224 L = 224 mL.

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