Question

-If 35.00 mL of 0.0200 M aqueous HCl is required to titrate 30.00 mL of an aqueous solution of NaOH to the equivalence point, the molarity of the NaOH solution is Blank 1 M.

-20.00 mL of 0.510 M NaOH is titrated with 0.740 M H₂SO₄. Blank 1 mL of H₂SO₄ are needed to reach the end point.

Answer 1

-If 35.00 mL of 0.0200 M aqueous HCl is required to titrate 30.00 mL of an aqueous solution of NaOH to the equivalence point, the molarity of the NaOH solution is ----------- M.

            HCl + NaOH = NaCl + H2O

We can use the following equation for finding the unknown morality or volume for the acid base titration.

v1s1 = v2s2

where, v stands for volume and s stands for the molarity(strength).

suffix 1 and 2 are for HCl and NaOH respectively.

So, v1 = 35 mL, s1= 0.02 M.

v2 = 30 mL, s2=?

s2 = v1s1/v2 = 35*0.02/30 M = 0.023 M

20.00 mL of 0.510 M NaOH is titrated with 0.740 M H₂SO₄. ----------- mL of H₂SO₄ are needed to reach the end point.

            For the 2^nd reaction,

2NaOH + H2SO4 = Na2SO4 + H2O

2 moles H⁺ is released by H2SO4, so the equation will be as following,

v1s1 = v2(2s2)

In this case, suffix 1 and 2 are for NaOH and H2SO4 respectively.

so, we have, v1=20 mL, s1=0.510 M, v2=? s2=0.74 M

v2 = 20*0.510/(2*0.74) = 6.89 mL of H2SO4 is required.