Calculate the e-m-f of the cell in V Pt - H2 -1-0 atm - CH3COOH 0-1 M - - NH3aq- 0-01 M - - H21-0 M -Pt KaCH3COOH-1-8- 10-5- KB-NH3-1-8- 10-5- if the value is -46-10-x- then what is the value of x-

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Nernst Equation

Calculate the...

Question

Calculate the e.m.f of the cell in V
$P t | H_{2} | (1.0 a t m) | C H_{3} C O O H (0.1 M) | | N H_{3} (a q, 0.01 M) | | H_{2} (1.0 M) | P t$
$K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}, K_{B^{'}} (N H_{3}) = 1.8 \times 10^{- 5}$ . if the value is $- 46 \times 10^{- x}$ , then what is the value of x?

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Solution

$C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}$

$C H_{3} C O O H$
$C H_{3} C O O^{-}$ $H^{+}$
Initial concentration (M)
$0.1$ 0
0
Final concentration (M)
$0.1 - x$ $x$ $x$
$\frac{x^{2}}{0.1 - x} = 1.8 \times 10^{- 5}$
$[H^{+}]_{L} = x = 1.3146 \times 10^{- 3}$
$N H_{3} + H_{2} O \to N H_{4}^{+} + O H^{-}$

$N H_{3}$ $N H_{4}^{+}$ $O H^{-}$
Initial concentration (M)
$0.01$ 0
0
Final concentration (M)
$0.01 - y$ $y$ $y$
$\frac{y^{2}}{0.01 - y} = 1.8 \times 10^{- 5}$
$y = 1.3146 \times 10^{- 3} = [O H^{-}]$
$[H^{+}]_{R} = \frac{10^{- 14}}{1.3146 \times 10^{- 3}} = 7.4536 \times 10^{- 12}$
$E = E^{0} - \frac{0.0592}{n} l o g (\frac{[H^{+}]^{2}}{P_{H 2}})_{L} (\frac{P_{H 2}}{[H^{+}]^{2}})_{R} = 0 - \frac{0.0592}{2} l o g (\frac{1.314 \times 10^{- 3}}{7.4536 \times 10^{- 11}})^{2} = - 0.46 V$

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C H_{3} C O O H = 1.8 \times 10^{- 5}

;

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N H_{4} O H = 1.8 \times 10^{- 5}

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]

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	$C H_{3} C O O H$	$C H_{3} C O O^{-}$	$H^{+}$
Initial concentration (M)	$0.1$	0	0
Final concentration (M)	$0.1 - x$	$x$	$x$

	$N H_{3}$	$N H_{4}^{+}$	$O H^{-}$
Initial concentration (M)	$0.01$	0	0
Final concentration (M)	$0.01 - y$	$y$	$y$

Calculate the e.m.f of the cell in VPt|H2|(1.0atm)|CH3COOH(0.1M)||NH3(aq,0.01M)||H2(1.0M)|PtKa(CH3COOH)=1.8×10−5,KB′(NH3)=1.8×10−5. if the value is −46×10−x, then what is the value of x?