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Otrada [13]
3 years ago
15

If the lab technician needs 30 liters of a 25% acid solution, how many liters of the 10% and the 30% acid solutions should she m

ix to get what she needs? Enter your answer. CHECK ANSWER
Chemistry
1 answer:
Mice21 [21]3 years ago
6 0

Answer:

7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.

Explanation:

Let the volume of 10% acid solution used to make the mixture = x L

So, the volume of 30% acid solution used to make the mixture = y L

Total volume of the mixture = <u>x + y = 30 L .................. (1) </u>

For 10% acid solution:

C₁ = 10% , V₁ = x L

For 30% acid solution :

C₂ = 30% , V₂ = y L

For the resultant solution of sulfuric acid:

C₃ = 25% , V₃ = 30 L

Using  

C₁V₁ + C₂V₂ = C₃V₃

10×x + 30×y = 25×30

So,  

<u>x + 3y = 75 .................. (2) </u>

Solving 1 and 2 we get,

<u>x = 7.5 L </u>

<u>y = 22.5 L</u>

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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

6 0
4 years ago
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