What is the pH of the solution?

Question

5.77 g of nitrous acid ( HNO2 ) is added to water to make a final solution of 444 ml.
What is the pH of the solution ?

Use Ka = 4.50 x 10-4 at 25.0oC.

Use the approximation that x << the initial concentration

Answer 1

MM HNO2 =47 g /mol

5.77/47 =0.123 mole

Molarity = 0.123 / 0.444 L=0.276 M

HNO2<> H+ + NO2-

At equilibrium
0.276-x.....x........x
4.5 10^-4 =(x)(x) / 0.276-x

x=0.0111 = concentration H+

pH=1.95

Answer 2

Get the Molarity of the 5.77g of HNO2, by dividing it by the molar mass of 47. Then divide it by the volume of .444 li, which gives a molarity of 0.276 M.

let x the acid concentration and using your assumption.
you get x^2 = (4.50 x 10^-4)* (0.276)
which is approximately 1.11 x 10^-2 or a

pH = 1.95. or with correction pH = 1.97

Answer 3

Fist, calculate the Molarity of HNO2:
M=moles/liters

5.77g=5.77/47.0=.123moles
M=.123/.444=.277M

HNO2<->H+ + NO2-
Initial: .277<->0 + 0
Change: -x <-> +x + x
_________________________________
Equilibrium: .277-x<-> x + x

Ka=concentration of products/reactants
4.5*10^-4=x^2/(.277-x)
then solve for x: x=3.45*10^-5

then to find the pH,
-log(3.45*10^-5)=4.46

Answer 4

HNO2 = (H+) +(NO2- )

and [H+] [NO2-]/[NO2H] = Ka

we search the molarity of NO2H MW= 14+2*16+1=47

5.77 in0.444ml in 1L c= 5.77/0444 =13.11g/l
The molarity is 13.11/47 =0.28 mole/l

lets [H+] [NO2-] =x^2 [NO2H] = c(1-x) and as x is small 1-x=1

so x^2 = Ka*c

log (1/x^2) = log 1/Ka-log c log (1/x^2) =2pH

pH = (pka-log c)/2= (3.347+0.553)/2 =1.9

pH=1.9