What is the pH for the solution made from 11.10 mL of 0.278 M HCl solution mixed with 70.90 mL of 0.1300 M ammonia solution ? Note that the pKa for NH4+ is 9.241
8.94
9.24
8.77
9.72
9.54
hey i found the working to this somewhere else and it was
its a buffer so
pH=pKa + log[base]/[acid]
=9.241 + log [70.9x.1300]/[11.1x0.278]
= 9.72
which makes sense but...
i also had that question on my assignment and got it wrong
the actual answer is 9.54
i cannot find a reason for this but as we were told not to use that eqn in chem then there must be another way to do it
To find the pH of the solution, we need to determine the concentration of the NH4+ and OH- ions in the solution, and then use the concentrations to calculate the pOH and convert it to pH.
First, we need to calculate the number of moles of HCl and ammonia in the solution.
For HCl:
Number of moles = Volume (in L) × Molarity
Number of moles = 11.10 mL × (1 L / 1000 mL) × 0.278 M
Number of moles = 0.003078 mol
For ammonia:
Number of moles = Volume (in L) × Molarity
Number of moles = 70.90 mL × (1 L / 1000 mL) × 0.1300 M
Number of moles = 0.009217 mol
Now, we need to determine the concentration of NH4+ and OH- ions.
The balanced equation for the reaction between HCl and ammonia is:
HCl + NH3 → NH4+ + Cl-
From the equation, we can see that 1 mole of HCl reacts with 1 mole of ammonia to form 1 mole of NH4+ ion.
Therefore, the concentration of NH4+ ions is equal to the number of moles of ammonia:
Concentration of NH4+ ions = Number of moles of ammonia / Total volume of solution (in L)
Concentration of NH4+ ions = 0.009217 mol / (11.10 mL + 70.90 mL) × (1 L / 1000 mL)
Concentration of NH4+ ions = 0.009217 mol / 0.082 L
Concentration of NH4+ ions = 0.1124 M
The concentration of OH- ions can be determined by applying the Kw expression (Kw = [H+][OH-]):
Kw = 10^-14 (at 25°C)
Since we know that the concentration of H+ and OH- ions are equal (assuming water is pure), we can calculate the concentration of OH- ions.
OH- concentration = Kw / [H+] concentration
OH- concentration = 10^-14 / [H+] concentration
To find [H+] concentration, we need to use the pKa value for NH4+.
pKa = -log10(Ka)
Rearranging the equation, we get:
Ka = 10^(-pKa)
Now, we can calculate [H+] concentration:
[H+] concentration = sqrt(Ka × [NH4+] concentration)
[H+] concentration = sqrt(10^(-pKa) × 0.1124 M)
Now that we have the concentration of [H+], we can calculate the pOH:
pOH = -log10(OH- concentration)
Finally, to determine the pH of the solution, we use the equation:
pH = 14 - pOH
Using the given pKa value for NH4+ (9.241) and the calculated concentrations, we find:
[H+] concentration = sqrt(10^(-9.241) × 0.1124 M)
[H+] concentration = 1.85 × 10^(-6) M
pOH = -log10(10^(-14) / (1.85 × 10^(-6)))
pOH = -log10(5.41 × 10^(-9))
pOH = 8.27
pH = 14 - 8.27
pH = 5.73
The pH of the solution made from the given volumes and concentrations is approximately 5.73.