Chemistry-pH,concentration

1.
(a)
[H⁺] = 0.0025 M
pH = -log(0.0025) = -2.60

(b)
[H⁺] = 0.15 x [20.0/(20.0 + 40.0)]= 0.05 M
pH = -log(0.05) = -1.30

(c)
CH3NH3 + H2O = CH3NH3⁺ + OH⁻

At eqm :
Let [CH3NH3⁺] = [OH⁺] = y M
Then [CH3NH2] = (0.10 - y) M

Kb = y²/(0.10 - y) = 4.2 x 10⁻⁴
y² + (4.2 x 10⁻⁴)y - (4.2 x 10⁻⁵) = 0
y = 6.27 x 10⁻³ or y = -6.69 x 10⁻³ (rejected)

pOH = -log(6.27 x 10⁻³) = 2.2
pH = 14 - 2.20 = 11.8

(d)
NH3 + H⁺ → NH4⁺
[NH3] = 0.200 x (35.0/60.0) - 0.100 x (25/60) = 0.075 M
[NH4⁺] = 0.100 x (25/60.0) =0.25/6 M

NH4⁺= NH3 + H⁺
pH = pKa - log([NH4⁺]/[NH3])
pH = -log(5.70 x 10⁻¹⁰) - log((0.25/6)/0.075) = 9.5


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2.
CO + 2H2 = CH3OH + heat

(a)
The equilibrium concentration of CO is decreased.
In the equation, there are 3 moles of gaseous molecules on the right, and 1mole on the left. When the totalpressure is increased, according to Le Chatelier's principle, the equilibriumposition shifts to the right to decrease the number of gaseous molecules.

(b)
The equilibrium concentration of CO is decreased.
The decrease in volumeleads to the increase in total pressure, and thus this causes the same effecton equilibrium position as in (a).

(c)
The equilibrium concentration of CO is increased.
The addition of heat favours the endothermic reaction, i.e. the backwardreaction.

(d)
The equilibrium concentration of CO is decreased.
According to Le Chatelier's principle, the addition of CO shifts the equilibriumposition to the right.

(e)
The equilibrium concentration of CO is decreased.
According to Le Chatelier's principle, the removal of CH3OH shiftsthe equilibrium position to the right.

1a, assume 100% dissociation, 0.0025M H+, pH=2.60
1b assume 100% dissociation, 0.15M.0.05M pH= 1.30
1c Kb=[congegate acid][OH-]/[base], neglect water selfdissociation
4.2x10^-4= (x)(x)/(0.1-x)
you calculate yourself and x is [OH-], pH=(-1)log(10^-14/x)
1d, orginally: 0.1x25/60M-->A, 0.2x35/60-->B
Ka=[H+][NH3]/[NH4+]=(A-x)(B-x)/x, you solve it to find [H+] and (-1)log[H+]

2a
CO+2HydrogenGas-->methanol
K=(methanol)/ (CO)(H2)^2
a) pressure increase, it is observed the 分母increase faster than 分子, equilibrium shift to product
if pressure is increased by increasing total pressure by adding inert gas without changing other parameter like volume, than it would be not changed
b)=pressure increase, same with a
c)pressure increase slightly, but if it is exothermic, shift to reactant, if endothermic, shift to product because this can help to cool
d)shift to product, to compensate the increase
e)shift to product to compensate loss
f)shift to product
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