Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06

5 0

3 years ago

Read 2 more answers

if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius

Lady_Fox [76]

The mass of ore required is 21 700 t.

r = 750 cm

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3} \pi (750 cm)^{3}$ = 1.767 × 10⁹ cm³

The density of lead is 11.34 g/cm³.

So mass of lead sphere = 1.767 × 10⁹ cm³ × $\frac{11.34 g}{1 cm^{3} }$ = 2.004 ×10¹⁰ g

2.004 ×10¹⁰ g × $\frac{1 kg}{1000 g}$ = 2.004 × 10⁷ kg

2.004 × 10⁷ kg × $\frac{1 t}{1000 kg}$ = 2.004 × 10⁴ t

92.5% efficiency means 92.5 t Pb per 100 t of ore.

Mass of ore = 2.004 × 10⁴ t Pb × $\frac{100 tore}{92.5 t Pb}$ = 2.17 × 10⁴ t ore = 21 700 t ore

6 0

3 years ago

Which of the following is not an example of a molecule

brilliants [131]

C is the correct answer

6 0

3 years ago

ether has a density of 0.736 g/cm^3 at 20 degrees C. what is the volume, in liters, of 225.0 grams of ether?

polet [3.4K]

D = m / V

0.736 = 225.0 / V

V = 225.0 / 0.736

V = 305.7 cm³

5 0

3 years ago