Question

I know this is more chemistry, but I am confused on what to do mathematically! HELP
What is the pH of a 0.085 M solution of nitrous acid (HNO2) that has a Ka of 4.5 × 10-4?

Answer 1

@Michele_Laino

Answer 2

first step: is nitrous a strong acid or weak acid?

Answer 3

a weak acid, correct?

Answer 4

good

we can now write out our acid dissociation equation

Answer 5

Ka = [H+][NO2-]/[HNO2]

we let x = the concentration of H+ giving us
let C = the original concentration

Ka = x * x / (C-x)

now we plug in our Ka and C values to solve for x

Answer 6

4.5*10^(-4) = x^2/(0.085 - x)

Answer 7

shortcut method: we can assume that the dissociated x is very small, so 0.085 - x is almost 0.085

Answer 8

4.5*10^(-4) = x^2/0.085

solve for x

Answer 9

I tried that before, but got a ridiculously small number. I get 3.825X10^-5 = x^2 and then the square root of that is around .006184

Answer 10

correct!

Answer 11

it's supposed to be small

next step is to find

-log(x)

Answer 12

where x = .006184

Answer 13

(remember, x is not the pH. x is the H+ concentration)

Answer 14

Ohhhh That's where I was getting confused. so when I do -log(.006184) I get about 2.21!

Answer 15

good! that's correct

Answer 16

Yay! Thank you so much!