We know that [OH⁻] * [H⁺] = 10⁻¹⁴
plugging the value of [H⁺]
[OH⁻] * 1.2 * 10⁻³ = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ * (10³/1.2)
[OH⁻] = 833.3 * 10⁻¹⁴
[OH⁻] = 8.33 * 10⁻¹²
The solution for this problem is:
Get into moles first. .0560 grams over 540.8 grams per mole = 1.04 x l0^-4 moles
Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq)
Ksp = (Sr++)^3(As04^-3)^2
(Sr++) = 3 X 1.04 x l0^-4= 3.11 x l0^-4
(As04^-3) = 2 x 1.04 x l0^-4= 2.07 x l0^-4
Ksp = (1.04 x l0^-4)^3 (2.07 x l0^-4)^2 which equals 4.82 x 10^-20
I can try.
Whats the question
Answer:
Explanation:
Kindly answer the way I have written.
