Final answer:
The pH of a 0.050 M NaCN solution is calculated using its conjugate acid's dissociation constant and the relationship between pH and pOH. The solution will be basic and the pH likely higher than 7, pointing to options A or D as potential answers.
Step-by-step explanation:
The pH of a solution made by mixing 0.050 mol of NaCN with enough water to make a liter of solution can be calculated using the dissociation constant (Ka) of its conjugate acid (HCN). Since NaCN is a salt of a weak acid HCN and a strong base NaOH, it will hydrolyze in water to form HCN and NaOH. The result is an increase in hydroxide ions (OH-), making the solution basic. The Ka of HCN is given as 4.9 x 10-10.
We can find the concentration of CN- in the solution, which is 0.050 M, as it is a 1:1 salt. The hydrolysis equation for CN- is:
CN- + H2O → HCN + OH-
The Kb for CN- can be calculated using the relation Kw = Ka * Kb, where Kw is the dissociation constant of water (1 x 10-14). Since Ka of HCN is 4.9 x 10-10, we can find Kb of CN-:
Kb = Kw / Ka = (1 x 10-14) / (4.9 x 10-10)
This yields a Kb of about 2.041 x 10-5. Using the formula:
OH- concentration = √(Kb * CN- concentration), we find the concentration of OH-.
Then, we can calculate the pOH using -log[OH-], and finally, use the relation pH + pOH = 14 to find the pH. Given the options provided, we expect the pH to be basic (greater than 7), which suggests the correct answer is likely either A or D.