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Chemistry

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svlad2 [7]3 years ago

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Marta_Voda [28]3 years ago

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A chemist titrates of a hypochlorous acid solution with solution at . Calculate the pH at equivalence. The of hypochlorous acid

const2013 [10]

The question is incomplete, here is the complete question:

A chemist titrates 110.0 mL of a 0.2412 M hypochlorous acid (HCIO) solution with 0.0613 M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of hypochlorous acid is 7.50. Round your answer to 2 decimal places

Answer: The pH of the solution is 10.09

Explanation:

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HClO$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.2412M\\V_1=110.0mL\\n_2=1\\M_2=0.0613M\\V_2=?mL$

Putting values in above equation, we get:

$1\times 0.2412\times 110.0=1\times 0.0613\times V_2\\\\V_2=\frac{1\times 0.2412\times 110.0}{1\times 0.0613}=432.8mL$

At equivalence, the number of moles of acid is equal to the number of moles of base. Also, the moles of salt which is NaClO will also be the same.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For HClO:

Molarity of HClO solution = 0.2412 M

Volume of solution = 110.0 mL

Putting values in equation 1, we get:

$0.2412M=\frac{\text{Moles of HClO}\times 1000}{110}\\\\\text{Moles of HClO}=\frac{(0.2412\times 110)}{1000}=0.026532mol$

For NaClO:

Moles of NaClO = 0.026532 moles

Volume of solution = [432.8 + 110] mL = 542.8 mL

Putting values in above equation, we get:

$\text{Molarity of NaClO}=\frac{0.026532\times 1000}{542.8}=0.0489M$

To calculate the pH of the solution, we use the equation:

$pH=7+\frac{1}{2}[pK_a+\log C]$

where,

$pK_a$ = negative logarithm of weak acid which is hypochlorous acid = 7.50

C = concentration of the salt = 0.0489 M

Putting values in above equation, we get:

$pH=7+\frac{1}{2}[7.50+\log (0.0489)]\\\\pH=7+3.09=10.09$

Hence, the pH of the solution is 10.09

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3 years ago

Radon-222 is a radioactive gas with a half-life of 3.82 days. How long would it take for fifteen-sixteenths of a sample of radon

Fittoniya [83]

Use the formula in terms of half life from the normal exponential functions

N(t) = N(0) (1/2) ^ (t/thalf) 

N(0) is the original quantity 

N(t) = quantity remaining at time t 

t is the time 


thalf is half life 

1/16 = (1/2)^(t/3.82) 

16 = 2^(t/3.82) 

4 = t/3.82 

t = 15.28 days

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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andriy [413]

Explanation:

A single-replacement reaction replaces one element for another in a compound.

A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.

A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.

Solubility rules are used to predict whether some double-replacement reactions will occur.

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Explanation:

hope this helps, I tried to give an easy definition :)

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