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Chemistry-All about pH,pOH,pka and pkb

I have picked out today to post about pH because i see that so many student trouble with it yet it's so easy. Come on now and take a ride on pH WITH ME.

The letter 'p' means the negative log to base 10 of.
CALCULATING PH
to compute this you need to know the concentration of the hydrogen or hydronium ion.
pH=-log(H+)
to calculate H+ if the pH is given we use
(H+)=10^-pH OR (H+)=antilog(-pH)

QUESTION 1. If the pH of a solution is 3.17, determing the concerntration of H+ in moles/L.
SOLUTION
pH=-log(H+)
pH=3.17,H+=?
But H+=10^-pH
H+=10^-3.17=

QUESTION 2. What is the pH of a 10^-4M Hcl solution.
SOLUTION
pH=-log(H+)
=-log(10^-4)
=4

CALCULATING pOH
pOH=-log(OH-)
to calculate OH- we use
(OH-)=10^-pOH
OR (OH-)=antilog(OH-)
Its just the same as the first

RELATIONSHIP BTW pH and pOH
At 25¤C the relation pH and pOH is given as
pH+pOH=14
if one of either pH or pOH is known the other can be found.

QUESTION 2.
A. If the H+ concerntration of a solution at 25¤C is 9.83*10^-5M, determine the concerntration of OH-. Use kw=1.00*10^-14.
B.If the OH- concerntration of a solution at 25¤C is 9.57*10^-9M, determine the pH. Use kw=1.00*10^-14.
Solution
a. H+=9.83*10^-5M
==>pH=-log(9.83*10^-5)
=-log9.83-log10^-5
solve this to find the value. But let me call the value x. From pH+pOH=14
pOH=14-x

b.Given that OH=9.57*10^-9 pOH=-log(9.57*10^-9)
=-log9.57-log10^-9
again let me call this value x.So that pH=14-x.

CALCULATING pKa and pKb
Ka and Kb is the equílibrium constant for ionization of the acid and base
pKa=-log(ka) and
pKb=-log(kb)

QUESTION 3. What is the pKa of acetic acid,if Ka for acetic acid is 1.78*10-5?
SOLUTION
pKa=-log(1.78*10-5)
=-(-4.75)
=4.75



This post first appeared on HOME WORK HELP(send All Questions To [email protected] And View Answer Here), please read the originial post: here

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