Calculate the pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) with 70.0 mL of 0.030 M NaCN.
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To calculate the pH of the solution resulting from the mixing of HCN(aq) and NaCN, we need to use the Henderson-Hasselbalch equation, which is:
pH = pKa + log([A-]/[HA])
First, we need to determine the concentrations of the acid (HA) and its conjugate base (A-).
The acid, in this case, is HCN(aq), and the base is CN-(aq) (the conjugate base of HCN).
Given:
Volume of HCN(aq) = 30.0 mL = 0.030 L
Concentration of HCN(aq) = 0.050 M
To find the amount of HCN(aq):
Amount of HCN(aq) = Volume x Concentration = 0.030 L x 0.050 M = 0.0015 moles
Next, let's determine the amount of CN-(aq) by using the concentration and volume of NaCN:
Volume of NaCN = 70.0 mL = 0.070 L
Concentration of NaCN = 0.030 M
Amount of CN-(aq) = Volume x Concentration = 0.070 L x 0.030 M = 0.0021 moles
Now, let's calculate the total amount of CN-(aq):
Total amount of CN-(aq) = Amount of CN-(aq) + Amount of CN-(aq) from HCN(aq)
= 0.0021 moles + 0.0015 moles
= 0.0036 moles
Since the stoichiometry of HCN(aq) and CN-(aq) is 1:1, the amount of CN-(aq) is the same as the amount of the conjugate base (A-).
Now, we need to calculate the ratio of [A-]/[HA]:
[A-]/[HA] = (0.0036 moles)/(0.0015 moles) = 2.4
Next, we need to determine the pKa value of HCN. The pKa of HCN is 9.21.
Now, let's substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= 9.21 + log(2.4)
Using a calculator, we find that log(2.4) is approximately 0.38.
Therefore,
pH = 9.21 + 0.38
= 9.59
The pH of the solution resulting from the mixing of 30.0 mL of 0.050 M HCN(aq) with 70.0 mL of 0.030 M NaCN is approximately 9.59.