Agbr has very low solubility in water silver ions form a...

AgBr has very low solubility in water. Silver ions form a complex ion with thiosulfate. c.o't''e'9 Ag+(aq) + 2S2o;-(aq) [Ag(S2 0 3 ) 2 ] 3 -(aq) If sodium thiosulfate is added to a saturated solution of AgBr in equilibrium with solid AgBr, how will the concentrations of free Ag+(aq) and Br-(aq) change? (A) Ag+(aq) decreases, Br-(aq) decreases (C) Ag+(aq) increases, Br-(aq) decreases (B) Ag+(aq) decreases, Br-(aq) increases (D) Ag+(aq) increases, Br-(aq) increases

Transcript

00:01 So in this question, silver ion, it forms complex ion.

00:08 So that is ag s2 .o3 -3 minus.

00:16 So whenever ag plus ion reacts with s2 -o -3 -2 -minus, that is the o -saltite, it forms the complex.

00:25 So according to the equation, they have given the equation ag plus atleus 2 s -2o3, 2 minus material.

00:34 S it is rise to vhe s2o33 minus rps.

00:42 Here, ks value is given 2 .9, it is given 2 .9 to 102.

00:46 40 % is solidity equal to be a mild.

00:49 Ks p of the silver bromide.

00:51 So what they have given, kst value also, they have given.

00:56 The ke .s .p.

00:57 Value 5 into 10 .10 ,000, minus 3.

01:00 What is the net equilibrium constant? net equilibrium constant for the following reaction that is ag dr surlade plus 2 s2 03 2 minus it used as to a g s2 03 taken twice 3 minus plus br minus so that's to calculate that also we have to calculate as well as a gdr in 0 .5 more of n2 s2o3 solution so much so motion so coming to the solution here.

01:35 So we know that ksp, ksp, that is concentration of product divided by concentration of reactant.

01:43 So whenever we see a equation as 2o3 minus aps, it is an equilibrium ag is 2o3 taken plus 3 minus.

01:58 So here, kf is equal to 2 .9 into 10 to the power of 30.

02:04 So, a, z, b, r solid.

02:08 It is in equilibrium a .g plus aqueous plus br minus acquies.

02:15 So k -u -s -p is equal to 5 into 10 to the power of minus 13...