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creativ13 [48]
1 year ago
14

A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to

report your answer with the correct number of significant figures
Chemistry
1 answer:
Gnom [1K]1 year ago
7 0

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

#SPJ4

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4 years ago
You are given the following three half-reactions:(1) Fe³⁺(aq) + e⁻ ⇄ Fe²⁺(aq) (2) Fe²⁺(aq) +2e⁻ ⇄ Fe(s) (3) Fe³⁺(aq) +3e⁻ ⇄ Fe(s
RideAnS [48]

The E°half-cell of the (3) is 0.33 volts .

Given,

(1)  Fe3+(aq) +e =Fe2+(aq)

(2) Fe2+(aq) +2e = Fe(s)

(3) Fe3+(aq) +3e = Fe(s)

We know ,

E° half-cell of (1) is 0.77 volts .

E° half cell of ( 2) is -0.44 volts .

By resolving or adding equation 1 and 2 we will get (3) ,

Thus , the value of the E° half cell of (3) is given by ,

E°cell = E° half-cell of (1 ) + E° half-cell of (2 ) = 0.77-0.44 = 0.33 volts

Hence the E° half-cell of ( 3 ) is 0.33 volts .

<h3>What is cell potential? </h3>

It is the difference between the electrode potentials of two half cells.

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