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3 years ago

Hydroxylamine is a weak molecular base with kb = 6.6 x 10-9. what is the ph of a 0.0500 m solution of hydroxylamine?

Chemistry

1 answer:

7nadin3 [17]3 years ago

4 0

Answer is: pH of hydroxylamine solution is 9,23.

Kb(NH₂OH) = 1,8·10⁻⁵.
c₀(NH₂OH) = 0,0500 M = 0,05 mol/L.
c(NH₂⁺) = c(OH⁻) = x.

c(NH₂OH) = 0,05 mol/L - x.
Kb = c(NH₂⁺) · c(OH⁻) / c(NH₂OH).

0,0000000066 = x² / (0,05 mol/L - x).

solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.
pOH = -log(0,000018 mol/L) = 4,74.
pH = 14 - 4,74 = 9,23.

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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

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Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

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$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

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