Item 7 Two 27.0 mL samples, one 0.100 molL1HCl and the other 0.100 molL1HF, were

Question

Item 7

Two 27.0 mL samples, one 0.100 molL1HCl and the other 0.100 molL1HF, were titrated with 0.200 molL1KOH.

Part A

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

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Part B

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

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Part C

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

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

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Part D

Predict which titration curve will have the lowest initial pH.

Predict which titration curve will have the lowest initial .

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Part E

Choose a rough sketch of each titration curve.

Choose a rough sketch of each titration curve.

Explanation / Answer

no of moles of HCl = molarity * volume in L

                              = 0.1*0.027 = 0.0027 moles

no of moles of KOH required = 0.0027 moles
volume of KOH                      = no of moles/molarity

                                              = 0.0027/0.2 = 0.0135 L = 13.5ml
The no of moles of HF are the same so the volume of KOH is the same = 13.5ml

at the equalent point

the strong acid (HCl) + strong base (KOH) >> PH= 7 the solution is neutral

Weak acid (HF) + strong base (KOH) >>> PH>7 the solution is basic

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