Two 23.0 mL samples, one 0.100 M HCl and the other 0.100 M HF, were titrated wit

Question

Two 23.0 mL samples, one 0.100 M HCl and the other 0.100 M HF, were titrated with 0.200 M KOH. Answer each of the following questions regarding these two titrations.

Part A What is the volume of added base at the equivalence point for HCl?

Part B What is the volume of added base at the equivalence point for HF?

Part C Predict whether the  at the equivalence point for each titration will be acidic, basic, or neutral.

Part D Predict which titration curve will have the lower initial pH.

neutral for HF, and basic for HCl neutral for HF, and acidic for HCl neutral for HCl, and acidic for HF neutral for HCl, and basic for HF neutral for both

Explanation / Answer

moles HCl = 0.0230 L x 0.100 M = 0.00230
moles KOH required = 0.00230
Volume KOH = 0.00230 / 0.200 M=0.0115 L => 11.5 mL

the moles of HF are the same so the volume of KOH is the same : 11.5 mL

at the equivalence point :
strong acid ( HCl) + strong base (KOH) = > pH = 7.0 the solution is neutral

weak acid (HF) + strong base (KOH) => pH > 7 the solution is basic

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