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
Hi Lucinda:
This is a buffer:
pH = pKa + log [base] / [acid]
pH = 9.241 + log [70.90 x 0.1300] /
[ 11.10 x 0.278]
= 9.72
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Zareh Darakjian, Ph.D.
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I must say, this solution has a bit of a pH-balancing act going on! Let's calculate it together. Since we have HCl and NH3 (which reacts to form NH4+), we need to compare the acid and base strengths to determine the pH. The pKa for NH4+ is 9.241, which means it acts as a weak acid. Now, let's crunch the numbers.
First, we find the moles of HCl:
(0.278 M) x (0.01110 L) = 0.0030838 moles
Then, we find the moles of NH4+:
(0.1300 M) x (0.07090 L) = 0.009217 moles
Now, we need to figure out the moles of NH3 that reacted to form NH4+:
0.009217 moles NH4+ x 1 mole NH3/1 mole NH4+ = 0.009217 moles NH3
To find the concentration of NH3:
0.009217 moles NH3 / (0.07090 L + 0.009217 L) = 0.1199 M NH3
Taking the negative logarithm (pH) of the Kb for NH3 gives us:
pH = 14 - pKa = 14 - 9.241 = 4.759
Since we have more moles of NH3 (base) than HCl (acid), the pH will be closer to basic. Therefore, we can conclude that the pH for this solution is NOT on the list of options you provided. Keep up the good work with those chemistry questions!
To determine the pH of the solution, we need to calculate the concentration of the ammonium ion (NH4+) and use the pKa value to find the pH.
Step 1: Calculate the moles of HCl:
moles of HCl = volume (in L) x concentration (in M)
moles of HCl = 0.01110 L x 0.278 M
moles of HCl = 0.0030878 mol
Step 2: Calculate the moles of ammonia:
moles of ammonia = volume (in L) x concentration (in M)
moles of ammonia = 0.07090 L x 0.1300 M
moles of ammonia = 0.009217 mol
Step 3: Determine the limiting reagent:
Both HCl and ammonia can react, however, HCl is in excess since its moles are greater than the moles of ammonia.
Step 4: Calculate the moles of ammonium ion:
moles of ammonium ion = moles of ammonia
moles of ammonium ion = 0.009217 mol
Step 5: Calculate the concentration of ammonium ion (NH4+):
volume of solution = volume of HCl + volume of ammonia
volume of solution = 0.01110 L + 0.07090 L
volume of solution = 0.082 L
concentration of ammonium ion = moles of ammonium ion / volume of solution
concentration of ammonium ion = 0.009217 mol / 0.082 L
concentration of ammonium ion = 0.1125 M
Step 6: Calculate the pOH using the Henderson-Hasselbalch equation:
pOH = pKa + log(base/acid)
base: concentration of ammonium ion (NH4+) = 0.1125 M
acid: concentration of ammonia (NH3) = 0.1300 M
pOH = 9.241 + log(0.1125/0.1300)
pOH = 9.241 + log(0.8654)
pOH = 9.241 + (-0.0631)
pOH = 9.177
Step 7: Calculate the pH:
pH = 14 - pOH
pH = 14 - 9.177
pH ≈ 4.823
Therefore, the pH of the solution is approximately 4.823.
To find the pH of the solution, we need to determine the concentration of the ammonium ion (NH4+), which is formed when ammonia (NH3) reacts with HCl.
First, let's calculate the amount of moles for each substance:
For HCl:
Moles of HCl = volume (in L) x concentration
Moles of HCl = 0.01110 L x 0.278 mol/L
Moles of HCl = 0.0030798 mol
For ammonia (NH3):
Moles of NH3 = volume (in L) x concentration
Moles of NH3 = 0.07090 L x 0.1300 mol/L
Moles of NH3 = 0.009217 mol
Since the stoichiometric ratio between HCl and NH4+ is 1:1 (1 mole of HCl reacts with 1 mole of NH3 to form 1 mole of NH4+), the moles of NH4+ formed will be equal to the moles of HCl used.
Now, let's calculate the concentration of NH4+:
Concentration of NH4+ = moles of NH4+ / total volume (in L)
Concentration of NH4+ = 0.0030798 mol / (0.01110 L + 0.07090 L)
Concentration of NH4+ = 0.0030798 mol / 0.0820 L
Concentration of NH4+ = 0.0375 mol/L
Now, let's calculate the pOH of the solution:
pOH = -log[OH-]
Since the NH4+ ion reacts with water to form OH-:
OH- concentration = NH4+ concentration
pOH = -log(0.0375) ≈ 1.43
Lastly, let's find the pH of the solution:
pH + pOH = 14
pH + 1.43 = 14
pH ≈ 12.57
Therefore, the pH of the solution is approximately 12.57.