20.0 mL of 0.06 M HCl (in a flask) is titrated with 0.04 M NaOH (in a burette). How many milliliters of NaOH needs to be used to reach the equivalence point?

A.0.03 mL

B.20.0 mL

C.20 mL

D.30 mL

2.Which of the following would help you identify a titration curve that involved a strong acid titrated by a weak base?

A.The pH at the equivalence point is lower than 7.

B.The pH at the equivalence point is higher than 7.

C.The titration curve begins at a higher pH and ends at a lower pH.

D.There is a rapid change in pH near the equivalence point (pH = 7).

0.0200 L X 0.06 mol/L HCl = 1.2X10^-3 mol HCl

Moles NaOH added to reach equivalence point = 1.2X10^-3 mol NaOH

Volume NaOH = 1.2X10^-3 mol / 0.0400 M NaOH = 0.030 L NaOH = 30 mL NaOH

D.There is a rapid change in pH near the equivalence point (pH = 7)
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AsiaWoerner AsiaWoerner · Answer 2 · 2019-05-12T11:55:31-04:00

Answer:

There is a rapid change in pH near the equivalence point (pH = 7).

Explanation:

At equivalence point the moles of base used is equal to moles of acid taken.

Let us calculate the moles of acid taken.

Moles = molarity X volume (L)

Given

molarity = 0.06 M

Volume =20.0 mL = 20 X 10⁻³ L

moles of HCl = 0.06 X 20 X 10⁻³= 1.2 X 10⁻³

moles of NaOH required = 1.2 X 10⁻³

\(volume=\frac{moles}{molarity}=\frac{1.2X10^{-3} }{0.04}=30X10^{-3}L=30mL\)

Thus volume of NaOH used at equivalence point = 30 mL

2) For pH titration curve (for example strong acid and weak base) we observe a rapid change in pH near equivalence point.