a
Kb of freaction is = 5.9x10^-11
let conc [OH] = x
5.9x10^-11 = x^2/(.2-x)
this gives x= 3.43514×10^-6and other value is negative so it is not considered
pOH= -log(3.43514×10^-6) = 5.48
pH= 14 - 5.48= 8.52
part b
K for the reaction is 1.547*10^-10
let conc [OH] = x
1.547*10^-10 = x^2/(.2-x)
we get x= 5.5623×10^-6
pOH = -log(5.5623×10^-6) = 5.26
pH = 14 -5.26 = 8.74
part c
NaNO3 is a salt of strong acid (HNO3) and strong base (NaOH). always such a reaction produces nuetral salt.
So pH of NaNO3 sol will always be 7 irrespective of its concentration.
Calculate all equilibrium concentrations and the pH of each of the following solutions. a. 0.20 M...
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Calculate the pH of each of the solutions
and the change in pH to 0.01 pH units caused by adding 10.0 mL of
2.93-M HCl to 570. mL of each of the following solutions.
pka is not given.
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Calculate the pH of each of the solutions and the change in pH
to 0.01 pH units caused by adding 10.0 mL of 2.52-M HCl to 340. mL
of each of the following solutions.
Change is defined as final minus initial, so if the pH drops
upon mixing the change is negative.
+ 18.55/24 points Previous Answers My Notes Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0...