Question #4a1f3

1 Answer
Apr 4, 2017

pOH=4.255*10^(-13)

Explanation:

From the concentration of HCl, we can calculate the pH using the following rule:
pH= - log[H_"3"O^"+"]
We obtain:
pH=-log[0.0235]=1.6289

Now we use the formula
pH+pOH =14
We can write that like this:
pOH =14-pH

Since pH=1.6289, we calculate pOH
pOH=14-1.6289=12.37

Now the [OH^"-"] can be calculated from the pOH by using this formula: [OH^"-"]=10^(-pOH)
We obtain: [OH^"-"]=10^(-12.37)=4.255*10^(-13)
For more information about pH, check [here]

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How is pH+pOH =14 established?
In water, the following (ionization) reaction occurs:
2 H_"2"O -> H_"3"O^"+" + OH^"-"

Therefore the equilibrium can be written like
K_"c"=([H_"3"O^"+"]*[OH^"-"])/[H_"2"O]
Since water is the solvent here, we do not take the concentration of water into consideration (is very big). We obtain the following expression:
K_"c"=[H_"3"O^"+"]*[OH^"-"]

The K_"c" in this equation represents a special number because we talk about the ionisation of water. Therefore we denote K_"c" as K_"w". The value of the K_"w" is measured at 25°C.
K_"w" (25°C) = 1*10^(-14)
This means we can say:
K_"c"=K_"w"=[H_"3"O^"+"]*[OH^"-"]=1*10^(-14)

To get from the [H_"3"O^"+"] (concentration H_"3"O^"+") to the pH, we use the following formula:
pH= - log[H_"3"O^"+"]
The same is true for the [OH^"-"], since we define pOH as
pOH=- log[OH^"-"]

Now if we take the Log from both sides of the K_"w" equation, we get:
log(1*10^(-14))=log([H_"3"O"]*[OH^-])
A mathematics rule tells us that multiplying inside the logarithm function is the same as adding these logarithms. Therefore we get
log(10^(-14))=log[H_"3"O"]+log[OH^-]

And now we can use the definitions of pOH and OH! We get:
log(10^(-14))=-pH - pOH
with log(10^(-14))=-14 we get the function
-pH-pOH =-14
Which is the same as
pH+pOH=14