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Nuetrik [128]
1 year ago
13

2. what volume of 0.0150-m hcl solution is required to titrate 150. ml of a 0.0100-m ca(oh)2 solution? (10 points)

Chemistry
1 answer:
laila [671]1 year ago
3 0

100ml volume of  0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.

Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,

    C1 V1 =C2 V2

where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1)  HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1)  Caoh2 solution.

While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give  in the expression we get the volume of the final concentration.

          V1=  C2 V2/ C1

               = 0.0100m . 150ml /0.0150M

               = 100ml

The volume of the hcl solution is 100ml.

To learn more about dilution formula please visit:

brainly.com/question/7208939

#SPJ4

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Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

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given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

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t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

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