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Naily [24]
2 years ago
13

In the titration of 50. 0 mL of 0. 400 M HCOOH with 0. 150 M LiOH, how many mL of LiOH are required to reach the equivalence poi

nt
Chemistry
1 answer:
mart [117]2 years ago
8 0

The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>
  • Molarity of acid, HCOOH (Ma) = 0.4 M
  • Volume of acid, HCOOH (Va) = 50 mL
  • Molarity of base, LiOH (Mb) = 0.15 M
  • Volume of base, LiOH (Vb) =?

MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

Learn more about titration:

brainly.com/question/14356286

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In the laboratory, a student dilutes 13.5 mL of a 11.6 M hydroiodic acid solution to a total volume of 100.0 mL. What is the con
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Explanation:

The dilution equation is presented as this: M_{s} V_{s} =M_{d} V_{d}.

·M= molarity (labeled as M)

·V= volume (labeled as L)

·s= stock solution (what you started with)

·d= diluted solution (what you have after)

Now that we know what each part of the formula symbolizes, we can plug in our data.

11.6M*13.5mL=M_{d} *100mL

We cannot leave it like this because the volumes must be in Liters, not milliliters. To convert this, we divide the milliliters by 1000.

13.5mL/1000=0.0135L     100mL/1000=0.1L

Now that we have the conversions, let's plug them into the equation.

11.6M*0.0135L=M_{d} *0.1L

The only thing that we need to do now is actually solving the answer.

M_{d} =\frac{11.6M*0.0135L}{0.1L}       M_{d} =1.566M

From the work shown above, the answer is 1.566M.

I hope this helps!! Pls mark brainliest :)

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__

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You are the medical examiner at the scene of a murder at a restaurant. There is a
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<h3>Further explanation</h3>

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