The ksp for silver(i) phosphate is 1.8 x 10 –18 . determine the silver ion concentration in a saturated solution of silver(i) phosphate

The silver ion concentration in saturated solution of silver (i) phosphate is calculated as follows.

write the equation for dissociation of silver (i) phosphate
that is Ag3PO4 (s) = 3Ag^+(aq) + PO4 ^3-(aq)

let the concentration of the ion be represented by x
ksp is therefore= (3x^3 )(x) = 1.8 x10 ^-18

27 x^3 (x) = 1.8 x10^-18
27x^4 = 1.8 x10^-18 divide both side 27
X^4 = 6.67 x10 ^-20

find the fourth root x = 1.6 x10 ^-5m

the concentration of silver ion is therefore = 3 x (1.6 x10^-5) = 4.8 x10^-5m

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-22T11:47:43+01:00

The concentration of silver ions is mathematically given as

C= 4.8 x10^-5m

What is the concentration of silver ions?

Generally, the equation for the Chemical Reaction is mathematically given as

Ag3PO4 (s) = 3Ag^+(aq) + PO4 ^3-(aq)

Therefore

Ksp= (3x^3 )(x)

Ksp= 1.8 x10^{-18}

Hence

27 x^3 (x) = 1.8 x10^-18

x^4 = 6.67 x10 ^{-20}

x = 1.6 x10^{-5}

In conclusion, the concentration of silver ion is

C=3x

C = 3* 1.6 x10 ^-5m

C= 4.8 x10^-5m