Question

D Question 22 1 pts Determine the molar solubility of BaF2 in 0.071 M Naf. It's Kop is 1.5 x 10% Our system can't handle your tiny answer, so multiply it by 1000 before entering.

Answer 1

The reaction between and NaF is

BaF

BaF2 + NaF BaF + NaF2

The concentration of NaF = 0.071 M

Solubility product of BaF2,

Ksp = 1.5 * 10-6

The dissociation of is

BaF

$\small BaF_2\rightarrow Ba^{2+}+2F^-$

Therefore concentration of $\small [F^-]$ = 0.071M

	$\small [BaF_2]$(s)	$\small [Ba^{2+}]$(aq)	$\small [F^-]$(aq)
Initial (I)	- - -	0	0.071
Change (C)	- - -	+x	+2x
Equilibrium (E)	- - -	x	0.071+2x

Solubilitypproduct of BaF2 is given by,

$\small K_{sp} =[Ba^{2+}][F^-]^2$

$\small 1.5*10^{-6}=[Ba^{2+}][F^-]^2$

$\small 1.5*10^{-6}=x*(0.071+2x)^2$

Since x is very small value, the equation becomes,

$\small 1.5*10^{-6}=x*(0.071)^2$

$\small x=\frac {1.5*10^{-6}} {(0.071)^2} =2.97*10^{-4}M$

That is the concentration of [Ba2+] = $\small 2.97*10^{-4}M$ . Therefore molar solubility of BaF2,

$\small s=2.97*10^{-4}M$

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