Question

Calculate the volume in mL of a 1.420 M NaOH solution required to titrate the following solutions:

(a) 25.00 mL of a 2.430 M HCI solution

(b) 25.00 mL of a 4.500 M H2SO4 solution

(c) 25.00 mL of a 1.500 M H3PO4 solution

Answer 1

=158.4ml

Step-by-step explanation:

Solution:

Chemical\; reaction:Chemicalreaction:

H_2SO_4(aq)\;+\;2NaOH(aq)\implies Na_2SO_4(aq)\;+\;H_2O(l)H

2

SO

4

(aq)+2NaOH(aq)⟹Na

2

SO

4

(aq)+H

2

O(l)

1) By using the acid-base equation:

M_1V_1=M_2V_2M

1

V

1

=M

2

V

2

4.5M*25ml=1.42M*V_24.5M∗25ml=1.42M∗V

2

V_2=\frac{(4.5M*25ml)}{1.42M}V

2

=

1.42M

(4.5M∗25ml)

Remember 1 mole H2SO4 is equivalent to 2 moles NaOH, that is Normality of H2SO4 = (Molarity x 2)

79.2*2ml=158ml\;of\;1.42M\;NaOH\;will\;be\;required.79.2∗2ml=158mlof1.42MNaOHwillberequired.

Answer:

V_2=158.4mlV

2

=158.4ml