Question

date:
name(s):

1. hydrogen cyanide has the chemical formula hcn. the compound

occurs as a colourless gas or a pale blue liquid. its solution in water is
called hydrocyanic acid. it reacts with water as per the reaction
below. what is the ph of 0.20 mol/l solution of the above acid (kₐ =
6.2 ×10⁻¹⁰)?
hcn(aq) + h₂o(l) ⇔ cn⁻(aq) + h₃o⁺(aq)
(answer + units)
~ 6 of 6 ~
sch₄u

Explanation:

Step1: Define the acid dissociation constant expression

For the reaction $\ce{HCN(aq) + H_{2}O(l)
ightleftharpoons CN^{-}(aq) + H_{3}O^{+}(aq)}$, the acid dissociation constant $K_{a}$ is given by $K_{a}=\frac{[\ce{CN^{-}}][\ce{H_{3}O^{+}}]}{[\ce{HCN}]}$. Let $x = [\ce{H_{3}O^{+}}] = [\ce{CN^{-}}]$ (since they are produced in a 1:1 ratio from the dissociation of HCN). The initial concentration of HCN is $0.20\ \text{mol/L}$, and at equilibrium, $[\ce{HCN}] = 0.20 - x$. Since $K_{a}$ is very small ($6.2\times 10^{-10}$), $x$ will be much smaller than $0.20$, so we can approximate $0.20 - x \approx 0.20$.

Step2: Substitute values into the $K_{a}$ expression

Substituting into the $K_{a}$ formula: $6.2\times 10^{-10}=\frac{x\cdot x}{0.20}$. This simplifies to $x^{2}=6.2\times 10^{-10}\times 0.20$.

Step3: Solve for $x$ (concentration of $\ce{H_{3}O^{+}}$)

Calculate $x^{2}$: $x^{2}=1.24\times 10^{-10}$. Then take the square root of both sides: $x = \sqrt{1.24\times 10^{-10}} \approx 1.11\times 10^{-5}\ \text{mol/L}$.

Step4: Calculate pH

The pH is defined as $\text{pH}=-\log_{10}([\ce{H_{3}O^{+}}])$. Substituting $x = 1.11\times 10^{-5}$: $\text{pH}=-\log_{10}(1.11\times 10^{-5}) \approx 4.95$.

Answer:

$4.95$ (no units for pH, but as per the problem's request for answer + units, since pH is dimensionless, we can write $4.95$ (pH is unitless, but if we consider the problem's note, we can present it as $4.95$ (unit: none, but typically pH has no units))