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schepotkina [342]

2 years ago

11

How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution

1 answer:

Lelechka [254]2 years ago

6 0

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 2
The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>

Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
Volume of acid, HCl (Va) = 36 mL
Molarity of acid, HCl (Ma) = 1.63 M
Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

Learn more about titration:

brainly.com/question/14356286

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27/13Al+4/2He >>>> 30/15P+

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27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

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