a) Equllibrium concentration= = = N 2 H 4 ( x ) × N 2 H 5 + O H − ( x ) N_2H_4(x)\times N_2H_5^+OH^-(x) N 2 H 4 ( x ) × N 2 H 5 + O H − ( x )
Equation for equilibrium constant
= = = [ N 2 H 6 + ] [ O H − ] N 2 H 4 [N_2H_6^+][OH^-]\over N_2H_4 N 2 H 4 [ N 2 H 6 + ] [ O H − ]
Substituting values in the equation yields;
x × x 0.01 − x x\times x\over0.01-x 0.01 − x x × x = 8.9 × 1 0 − 16 =8.9\times 10^{-16} = 8.9 × 1 0 − 16
= x 2 + ( 8.9 × 1 0 − 16 × 0.01 M ) 1 2 =x^2+(8.9\times10^{-16}\times0.01M)^{1\over 2} = x 2 + ( 8.9 × 1 0 − 16 × 0.01 M ) 2 1
Using the quadratic formula;
a = 1 b = 8.6 × 1 0 − 7 c = − 8.9 × 1 0 − 16 a=1
b=8.6\times 10^{-7}
c=-8.9\times 10^{-16} a = 1 b = 8.6 × 1 0 − 7 c = − 8.9 × 1 0 − 16
− b − + -b \stackrel{+}{-} − b − + b 2 − 4 a c 2 a \sqrt {{b^2-4ac}} \over2a 2 a b 2 − 4 a c
= = = − 8.6 × 1 0 − 7 − + -8.6\times 10^{-7} \overset{+}{-} − 8.6 × 1 0 − 7 − + ( 8.6 × 1 0 − 7 ) 2 − 4 ( 1 ) ( 8.9 × 1 0 − 16 2 ( 1 ) \sqrt{(8.6\times10^{-7})^2-4(1)(8.9\times 10^{-16}} \over 2(1) 2 ( 1 ) ( 8.6 × 1 0 − 7 ) 2 − 4 ( 1 ) ( 8.9 × 1 0 − 16
= − 1.699 × 1 0 − 6 M =-1.699\times 10^{-6}M = − 1.699 × 1 0 − 6 M
b) p H = − l o g K b × C o pH=-log\sqrt{K_b\times C_o} p H = − l o g K b × C o
= − l o g ( K b × C o ) 1 2 =-log (K_b\times C_o)^{1\over2} = − l o g ( K b × C o ) 2 1
= − l o g ( 8.9 × 1 0 − 16 × 0.01 M ) 1 2 =-log (8.9\times10^{-16}\times 0.01M)^{1\over2} = − l o g ( 8.9 × 1 0 − 16 × 0.01 M ) 2 1
= − l o g 2.983 × 1 0 − 9 =-log 2.983\times 10^{-9} = − l o g 2.983 × 1 0 − 9
p H = 8.5 pH=8.5 p H = 8.5
Comments