Answer:
THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.
Explanation:
Using
Ca VA / Cb Vb = Na / Nb
Ca = 0.0500 M
Va = 35 mL
Cb = 0.0200 M
Vb = unknown
Na = 2
Nb = 1
Equation for the reaction:
Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O
So therefore, we make Vb the subject of the equation and solve for it
Vb = Ca Va Nb / Cb Na
Vb = 0.0500 * 35 * 1 / 0.0200 * 2
Vb = 1.75 / 0.04
Vb = 43.75 mL
The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL
The volume of 0.0200 M calcium hydroxide that would be required to neutralize 35.00 mL of 0.0500 M nitric acid will be 43.75 mL
From the equation of the reaction:
[tex]Ca(OH)_2 + 2 HNO_3 ---> Ca(NO_3)_2 + 2 H_2O[/tex]
The mole ratio of calcium hydroxide to nitric acid is 1:2.
Mole of nitric acid = molarity x volume
= 0.0500 x 35.00/1000
= 0.00175
Thus, the mole of the calcium hydroxide can be calculated as:
0.00175/2 = 0.000875
Volume of calcium hydroxide = mole/molarity
= 0.000875/0.0200
= 0.04375 L
= 43.75 mL
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