10.0 mL of a 0.300M NH3 solution is titrated with a 0.100M HCl solution. Calculate the pH after the following additions of HCl.
0.00mL
10.00mL
30.00mL
40.00mL
To calculate the pH after each addition of HCl, we need to consider the reaction between NH3 and HCl and the resulting change in concentration of NH3 and NH4+.
The balanced equation for the reaction is:
NH3 + HCl → NH4+ + Cl-
First, let's calculate the initial concentration of NH3 in the 10.0 mL solution:
moles of NH3 = volume (L) * concentration (M)
moles of NH3 = 10.0 mL * (1 L / 1000 mL) * 0.300 M
moles of NH3 = 0.003 moles
Since the stoichiometry of the balanced equation is 1:1 between NH3 and NH4+, we can conclude that the initial concentration of NH4+ is also 0.003 M.
Now, let's calculate the pH after each addition of HCl:
1) 0.00 mL:
At this point, no HCl has been added, so the concentration of NH3 and NH4+ remains at 0.003 M. We can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([NH4+]/[NH3])
pH = pKa + log(0.003/0.003)
pH = pKa
As we do not have the pKa value, we cannot calculate the exact pH. However, we know that NH3 is a weak base and NH4+ is its conjugate acid. Therefore, the expected pH will be slightly acidic.
2) 10.00 mL:
Since 10.00 mL of 0.100 M HCl is added to the 10.0 mL of NH3 solution, we need to calculate the new concentrations of NH3 and NH4+ after the reaction occurs:
moles of HCl = volume (L) * concentration (M)
moles of HCl = 10.00 mL * (1 L / 1000 mL) * 0.100 M
moles of HCl = 0.001 moles
According to the stoichiometry of the balanced equation, this reaction consumes 0.001 moles of NH3 and produces 0.001 moles of NH4+. Therefore, the moles of NH3 and NH4+ remaining in the solution are:
moles of NH3 = 0.003 moles - 0.001 moles = 0.002 moles
moles of NH4+ = 0.003 moles + 0.001 moles = 0.004 moles
Now, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([NH4+]/[NH3])
pH = pKa + log(0.004/0.002)
pH = pKa + log(2)
Again, without the pKa value, we cannot calculate the exact pH. However, since HCl is a strong acid and NH4+ is its conjugate base, we expect the pH to be acidic, lower than the pH obtained in the first step.
3) 30.00 mL:
Using the same approach, we can calculate the new concentrations of NH3 and NH4+ after adding 30.00 mL of 0.100 M HCl:
moles of HCl = 30.00 mL * (1 L / 1000 mL) * 0.100 M
moles of HCl = 0.003 moles
moles of NH3 = 0.003 moles - 0.003 moles = 0 moles
moles of NH4+ = 0.004 moles + 0.003 moles = 0.007 moles
Now, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([NH4+]/[NH3])
pH = pKa + log(0.007/0)
pH = undefined
In this case, because the concentration of NH3 becomes zero, the logarithm term becomes undefined. Therefore, the pH cannot be calculated.
4) 40.00 mL:
Following the same procedure:
moles of HCl = 40.00 mL * (1 L / 1000 mL) * 0.100 M
moles of HCl = 0.004 moles
moles of NH3 = 0 moles - 0.004 moles = -0.004 moles (consumed completely)
moles of NH4+ = 0.007 moles + 0.004 moles = 0.011 moles
Using the Henderson-Hasselbalch equation:
pH = pKa + log([NH4+]/[NH3])
pH = pKa + log(0.011/0)
pH = undefined
Again, because the concentration of NH3 becomes zero, the pH cannot be calculated.
In summary:
- At 0.00 mL and 10.00 mL of HCl, we expect a slightly acidic pH.
- At 30.00 mL and 40.00 mL of HCl, the pH cannot be calculated due to complete consumption of NH3.
I will do the third one.
NH4OH: .010l*.3=.0030 moles
HCl: .030*.1=.003 moles
they are all react, no excess acid or base. So, pH=7
the fourth one:
moles base: .0030
moles acid: .040*.1=.004, or an excess of acid of .001moles
volume+ 10ml+40ml=.050 liters
concen acid: .001/.050=.02M
pH= -log(.02)= 1.70