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Consider the following electrochemical cell at standard condition. 
$Au(s)$|$QH_2,Q$| $NH_4X(0.01 M)$ | | $Ag^+ (1M)$|$Ag(s)$ $E_{cell}$ = +0.4 V 
The couple $QH_2/Q$ represents quinhydrone electrode, the half cell reaction is given below : 

[Given: $E_{Ag^+/Ag}^ \circ$ = +0.8 V and $\frac{2.303RT}{F}$ = 0.06 V ]
The $pK_b$ value of the ammonium halide salt ($NH_4X$) used here is ________. (nearest integer)

To find the pKb value of the ammonium halide salt (NH4X), follow these steps:

1. Calculate the standard cell potential (E°cell):

The standard reduction potential for the cathode (Ag+/Ag) is 0.8 V and for the anode (Q/QH2) is 0.7 V.
E°cell = E°cathode - E°anode
E°cell = 0.8 V - 0.7 V = 0.1 V

2. Determine the pH using the Nernst Equation:

The overall cell reaction involves 2 electrons (n=2):
QH2 + 2Ag+ → Q + 2Ag + 2H+
Ecell = E°cell - (0.06 / n) * log([H+]2 * [Q] / [Ag+]2 * [QH2])

In a quinhydrone electrode, [Q] = [QH2]. Given [Ag+] = 1 M:
Ecell = E°cell - (0.06 / 2) * log([H+]2)
0.4 = 0.1 - 0.03 * 2 * log([H+])
0.4 = 0.1 - 0.06 * log([H+])

Since pH = -log([H+]):
0.4 = 0.1 + 0.06 * pH
0.3 = 0.06 * pH
pH = 5

3. Calculate the pKb for the salt NH4X:

NH4X is a salt of a weak base (NH3) and a strong acid (HX). The formula for the pH of such a salt is:
pH = 1/2 * (pKw - pKb - log C)

Given pKw = 14 and concentration C = 0.01 M (or 10-2 M):
5 = 1/2 * (14 - pKb - log(10-2))
5 = 1/2 * (14 - pKb - (-2))
10 = 14 - pKb + 2
10 = 16 - pKb
pKb = 6

The pKb value of the ammonium halide salt (NH4X) used here is 6.

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Important Questions from Physical Chemistry

  1. $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$ 
    Consider the above reaction, what mass of $CaCl_2$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of $CaCO_3$ ? 
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  5. If equal volumes of $AB_2$ and $XY$ (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $AY_2$ at 300 K ? 
    (Given $K_{sp}$ (at 300 K) for $AY_2=5.2 \times 10^{-7}$)

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