Question

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).

Answer 1

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)
hope it helps

Answer 2

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

Step-by-step 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=(moles)/(molarity)=(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.