Hi
"Ka for HCN is 4.9 x 10^-10. What is the pH of a 0.068 M aqueous solution of sodium cyanide?"
How do I do this question?
Thank you very much
This is a hydrolysis question. The CN^- ion is base and it reacts with water as follows:
CN^- + HOH ==> HCN + OH^-
Kb = (Kw/Ka) = (HCN)(OH^-)/(CN^-)
X = HCN
X = OH
CN is given in the problem. Kw you know, Ka is given in the problem. Solve for X, convert OH to pOH, then to pH.
To find the pH of a solution of sodium cyanide (NaCN), we need to consider the hydrolysis reaction of the cyanide ion (CN-) with water.
The equation for the hydrolysis reaction of CN- with water is as follows:
CN- (aq) + H2O(l) ⇌ HCN(aq) + OH- (aq)
Since we are given the Ka value for HCN, we can use it to calculate the concentration of OH- in the solution. The Ka value can be related to the concentration of the products (HCN and OH-) over the concentration of the reactants (CN-) using the equation:
Ka = [HCN] [OH-] / [CN-]
Since HCN is a weak acid, we can assume that [CN-] is equal to the initial concentration of NaCN, which is 0.068 M.
Now let's solve for [OH-]:
[OH-] = (Ka * [CN-]) / [HCN]
Plug in the given values:
[OH-] = (4.9 x 10^-10) * (0.068) / (1)
[OH-] = 3.332 x 10^-11 M
Now, we can use the fact that in pure water, the concentration of H+ ions and the concentration of OH- ions are equal. Thus, the concentration of H+ in the solution is also 3.332 x 10^-11 M.
To find the pH, we can use the equation:
pH = -log[H+]
Plug in the value for [H+]:
pH = -log(3.332 x 10^-11)
pH = 10.476
Therefore, the pH of a 0.068 M aqueous solution of sodium cyanide is approximately 10.476.
To solve this question, you need to understand the relationship between Ka, pH, and the concentration of a weak acid or base.
1. Determine the dissociation equation: Start by writing the balanced chemical equation for the dissociation of sodium cyanide (NaCN) in water. NaCN dissociates into sodium ions (Na+) and cyanide ions (CN-).
NaCN(aq) ⇌ Na+(aq) + CN-(aq)
2. Identify the weak acid or base: In this case, we are interested in the behavior of CN-. The CN- ion can act as a weak base because it has the potential to accept a proton (H+ ion) and form hydrogen cyanide (HCN) in water.
3. Write the equilibrium expression: The equilibrium expression for the dissociation of CN- can be written as:
Ka = [HCN][OH-] / [CN-]
However, if we assume the concentration of OH- to be negligible (since we are dealing with a weak base), we can simplify the expression to:
Ka = [HCN] / [CN-]
4. Write the equilibrium concentrations: We are given that the concentration of NaCN is 0.068 M. Since NaCN will completely dissociate into Na+ and CN- ions, we can assume the concentration of CN- is also 0.068 M.
5. Substitute values and solve for [HCN]: Plug the given values into the simplified equilibrium expression and solve for [HCN].
4.9 x 10^-10 = [HCN] / 0.068
[HCN] = 4.9 x 10^-10 x 0.068
[HCN] ≈ 3.332 x 10^-11 M
6. Calculate the pOH: Since we have [OH-] = 0, the pOH can be determined as:
pOH = -log10 [OH-] = -log10(0) = undefined
7. Calculate the pH: Since pH + pOH = 14, we can calculate the pH as:
pH = 14 - pOH
Therefore, the pH of the 0.068 M aqueous solution of sodium cyanide is approximately 14.
Please note that the calculated pH value of 14 indicates that the solution is highly basic, which is expected for a solution of a strong base like cyanide ions.