what is the maximum possible concentration of Ni2+ ions in a solution containing 0.15 M HCl and 0.1 M H2S. given that Ksp(NiS)= 2*10^-21 and [S2-]=4*10^-21.

Dear Parth,

find the Ka for H2S acid as l.l x l0-7 in pure water.

Ka expression l.l x l0-7 = (H+)^2(S=) over (H2S)

but in this solution the pH is 3.4, so next find the H+ concentration.

H+) – antilog of -pH or antilog of -3.4 which equals 3.98 x l0-4 moles per liter (H+) conc.

Plug this value into your Ka expression for the weak acid equilibrium and solve for maximum concentration allowed for the Sulfide ion.

Ka = l.l x l0-7 = (3.98 x l0-4)(S=) over .13M H2S)

l.43 x l0-8 = 3.98 x l0-4(S=)

so (S=) = 3.59 x l0-5 allowable S= concentration that can combine with Ni ion.

Now plug this number into the Ksp expression for NiS

Ksp = l.3 x l0-25 = (Ni++)(S=) = (Ni++)(3.98 x l0-5)

so (Ni++) = l.3 x l0-25 divided by 3.98 x l0-5 which equals 3.62 x l0-21st moles/liter Ni++ max concentration

Best Of luck

Plz Approve the answer...!!!!

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