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kaheart [24]
2 years ago
5

Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl after adding 50 mL of 0.1000 M NaOH:

Chemistry
1 answer:
Nana76 [90]2 years ago
5 0

the pH during the titration of 40.00 mL of 0.1000 M HCl after adding 50 mL of 0.1000 M NaOH is pH = 13.39

NaOH + HCl → NaCl + H2O

40.00 ml 0.5M NaOH solution

mol NaOH = 40 ml / 1000 ml / L * 0.5 mol / L = 0.02 mol NaOH

25 mL 0.1 M HCl

mol HCl = 25 mL / 1000 mL / L * 0.1 mol / L = 0, 0025 mol

This reacts with 0.0025 mol NaOH

[NaOH] = 0.269 M

[OH-] = 0.269 M

pOH = - log 0.269

pOH = 13.43

You titrate with 30 mL 0.100 M HCl

Mol HCl = 30 mL / 1000 mL/L * 0.100 mol /L = 0.003 mol HCl

This will react with 0.003 mol NaOH

Mol unreacted NaOH = 0.02 mol - 0.003 mol = 0.017 mol NaOH in 40mL + 30 mL = 70 mL solution

Molarity of NaOH solution = 0.017 mol / 70 mL * 1000 mL/L = 0.243 M

[OH-] = 0.243 M

pOH = - log 0.243

pOH = 0.61

pH = 13.39

40 mL

Titration is defined as "the process of determining the amount of substance A by adding a measured amount of substance B, which is a titrator that reacts until an accurate chemical equivalence point is reached."

Learn more about titration here: brainly.com/question/186765

#SPJ4

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What is the pH of a solution made by mixing 200 mL of 0.025M HCl and 150 mL of 0.050 M HCl?
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Answer:

The answer to your question is pH = 1.45  

Explanation:

Data

pH = ?

Volume 1 = 200 ml

[HCl] 1 = 0.025 M

Volume 2 = 150 ml

[HCl] 2 = 0.050 M

Process

1.- Calculate the number of moles of each solution

Solution 1

                Molarity = moles / volume

-Solve for moles

                moles = 0.025 x 0.2

result

               moles = 0.005

Solution 2

               moles = 0.050 x 0.15

-result

                moles = 0.0075

2.- Sum up the number of moles

Total moles = 0.005 + 0.0075

                   = 0.0125

3.- Sum up the volume

total volume = 200 + 150

                     350 ml or 0.35 l

4.- Calculate the final concentration

Molarity = 0.0125 / 0.35

              = 0.0357

5.- Calculate the pH

pH = -log [H⁺]

-Substitution

pH = -log[0.0357]

-Result

pH = 1.45    

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