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Zeeshan Sama
Asked: 2022-11-07T03:19:52+05:30 In:

`500ml` of `0.1MKCI, 200ml` of `0.01M NaNO_(3)` and `500ml` of `0.1M AgNO_(3)` was mixed. The molarity of `K^(+), Ag^(+),CI^(-),Na^(+),NO_(3)^(-)` in the solution would be
A. `[K^(+)] = 0.04, [Ag^(+)] = 0.04, [Na^(+)] = 0.002`
B. `[K^(+)] = 0.04, [Na^(+)] = 0.00166, [NO_(3)^(-)] = 0.04333`
C. `[K^(+)] = 0.04, [Ag^(+)] = 0.05, [Na^(+)] = 0.0025`
D. `[K^(+)] = 0.05, [Na^(+)] = 0.0025, [Cl^(-)] = 0.05, [NO_(3)^(-)] = 0.0525`

`500ml` of `0.1MKCI, 200ml` of `0.01M NaNO_(3)` and `500ml` of `0.1M AgNO_(3)` was mixed. The molarity of `K^(+), Ag^(+),CI^(-),Na^(+),NO_(3)^(-)` in the solution would be
A. `[K^(+)] = 0.04, [Ag^(+)] = 0.04, [Na^(+)] = 0.002`
B. `[K^(+)] = 0.04, [Na^(+)] = 0.00166, [NO_(3)^(-)] = 0.04333`
C. `[K^(+)] = 0.04, [Ag^(+)] = 0.05, [Na^(+)] = 0.0025`
D. `[K^(+)] = 0.05, [Na^(+)] = 0.0025, [Cl^(-)] = 0.05, [NO_(3)^(-)] = 0.0525`
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1 Answer

  1. Correct Answer – B
    `[K^(+)] = (0.1 xx 500)/(500 +200 +500) = (50)/(1200) = 0.04167M`
    `[Na^(+)] = (0.01 xx 200)/(500 +200 +500) = (2)/(!200) = 0.00167M`
    `[NO_(3)^(-)] = (200 xx 0.01 xx 500 xx 0.1)/(500 +200 +500) = (52)/(1200) = 0.0433M`
    Since 50 milli moles of `Cl^(-)` are mixed with 50 milli moles of `Ag^(+)` ions, total `Ag^(+)` & `Cl^(-)` get precipitated as `AgCl`.
    Hence the solution does not contain
    `Ag^(+)` & `Cl^(-)` ions

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