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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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