What is the molarity of a solution of calcium hydroxide, Ca(OH)_2, if 353 mL is

Question

What is the molarity of a solution of calcium hydroxide, Ca(OH)_2, if 353 mL is neutralized with 45.96 mL of a 0.425 M HCI? Potassium acid phthalate, KHC_8 H_4 O_4, is a crystalline solid that is available in a state of high purity, making it an excellent choice as a standard acid. What is the concentration of a NaOH solution, if it takes 43.00 mL to neutralize a 0.0584 g sample of the acid? The crystalline solid dissociates in water as follows: KHC_8 H_4 O_4 (aq) rightarrow K^+ (aq) + H^+ (aq) + C_8 H_4 O_4^2- (aq)

Explanation / Answer

1. Applying equivalent concept

milliequivalent of Ca(OH)2 = milliequivalent of HCl

Milliequivalent=molarity ×volume×n-factor

Milliequivalent of Ca(OH)2= M×353×2

Milliequivalent of HCl=45.96×0.425

Equating both we get

M=0.13 molar

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