At 298 {\K} the molar solubility of {\Cd}( {\OH})_2 in 0.1 {\M} {\KOH} solution is x \times 10^{-y}. The values of x and y are respectively (at 298 {\K}, {\K}_{ { sp}} of {\Cd}( {\OH})_2=2.5 \times 10^{-14}) [TG EAPCET 14 May 2023 Shift 2]

TS EAMCET Engineering 14 May 2023 Shift 2 Solved Paper

Section: Chemistry

Question : 129 of 160

Marks: +1, -0

At 298K the molar solubility of Cd(OH)2 in 0.1M‌KOH solution is x×10−y. The values of x and y are respectively (at 298K,K sp of Cd(OH)2=2.5×10−14)

[TG EAPCET 14 May 2023 Shift 2]

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