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Volgvan
2 years ago
13

When a 27.7 mL sample of a 0.400 M aqueous hypochlorous acid solution is titrated with a 0.335 M aqueous potassium hydroxide sol

ution, what is the pH at the midpoint in the titration
Chemistry
1 answer:
Harlamova29_29 [7]2 years ago
4 0

With a 27.7 mL sample of a 0.400 M aqueous hypochlorous acid solution  titrated with a 0.335 M aqueous potassium hydroxide solution, the pH value is mathematically given as

pH=12.90

<h3>What is the pH at the midpoint in the titration?</h3>

Moles of HClO=0.329*21.5

Moles of HClO=7.0735

and

Moles of KOH=0.456*23.3

Moles of KOH=10.6248

Hence

The net mole of KOH=10.6248-7.0735

The  net mole of KOH=3.5513

Volume of solution=21.5+23.3

Volume of solution=44.8ml

Generally, the equation for the  Concentration of KOH is mathematically given as

CKOH=mole/volume

Therefore

CKOH=3.5513/44.8ml

CKOH=0.07927M

In conclusion,

pOH=-log{OH-}

pOH=-log{0.07927}

pOH=1.10

pH=14-110

pH=12.90

Read more about pH value

brainly.com/question/15845370

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Answer:

607 ppm

Explanation:

In this case we can start with the <u>ppm formula</u>:

ppm=\frac{mg~of~solute}{Litters~of~solution}

If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of F^-, because the molarity formula is:

M=\frac{mol}{L}

In other words:

0.0320~M=\frac{mol}{1~L}

mol=0.032~M*1~L=0.032~mol

1~L~of~Solution=0.0320~mol~of~solute

If we use the <u>atomic mass</u> of F  (19 g/mol) we can convert from mol to g:

0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g

Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

0.607~g\frac{1000~mg}{1~g}=607~mg

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ppm=\frac{607~mg}{1~L}=~607~ppm

<u>We will have a concentration of 607 ppm.</u>

I hope it helps!

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Explanation:

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Unknown:

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