Question

Calculate [H3O+] in the following solutions.

1) 0.100 M NaNO2 and 5.00×10−2 M HNO2
2) 5.20×10−2 M HCl and 7.62×10−2 M NaC2H3O2


please help thx

Answer 1

I'm not going to work four problems this long for you. I'll do #1 and get you started on #2.

NaNO2 is determined by hydrolysis of the NO2^- and HNO2 is a weak acid.
................NO2^- + HOH ==>HNO2 + OH^-
I...............0.1...............................0...........0
C...............-x................................x............x
E.............0.1-x..............................x............x

Kb for NO2^- = (Kw/Ka for HNO2) = (x)(x)/(0.1-x)
Substitute and solve for x = (OH^-) and convert to (H3O^+)

For HNO2, that is a weak acid and
...............HNO2 ==> H^+ + NO2^-
I..............0.05............0.........0
C.............-x................x.........x
E..........0.05-x.............x.........x

Write the Ka expession for HNO2, plug in the E line and solve for x = H^+ = H3O^+

The object, I think, for having you do these two (in #1) is to show you that the salt is basic but the acid is acidic.

#2. HCl is a strong acid; therefore, (H^+) = (HCl) The acidity of the NaC2H3O2 (sodium acetate) is determined by the hydrolysis of the acaetate ion. That's done exactly the same way as the nitrite ion in #1.

Post your work if you get stuck.

Answer 2

I've calculate part 1)

................NO2^- + HOH ==>HNO2 + OH^-
I...............0.1...............................0...........0
C...............-x................................x............x
E.............0.1-x..............................x............x

Kb for NO2^- = (Kw/Ka for HNO2) = (x)(x)/(0.1-x)
4.5x10^-4 = (x)(x)/(0.1-x)
x = 6.75x10^-3 M
x = [OH-]
-log [OH-]
pH= 14- (-log [OH-])
pH =11.8


For HNO2, that is a weak acid and
...............HNO2 ==> H^+ + NO2^-
I..............0.05............0.........0
C.............-x................x.........x
E..........0.05-x.............x.........x

4.5x10^-4 = x^2 / 0.05-x
x = 4.52x10^-3
pH = 2.3

so which is the [H3O+] ? FOR QUESTION #1