Well, let me calculate that for you, but please don't hold your breath, as my calculations might leave you a little light-headed!
To determine the pH of the solution, we need to find the concentration of the resulting NH4+ (ammonium ion) and calculate the pOH based on that. Then we can subtract the pOH from 14 to get the pH.
First, let's find the number of moles of HCl and ammonia used:
moles of HCl = volume (L) x concentration (M)
= 11.10 mL x 0.278 M
= 0.0030878 mol
moles of ammonia = volume (L) x concentration (M)
= 70.90 mL x 0.1300 M
= 0.009217 mol
Now, let's calculate the number of moles of NH4+ formed by reacting HCl and ammonia:
1 mol of HCl reacts with 1 mol of NH3 to produce 1 mol of NH4+
So, the number of moles of NH4+ formed = 0.0030878 mol (since NH3 and HCl react in a 1:1 ratio)
Now, let's determine the concentration of NH4+:
concentration (M) = moles / volume (L)
= 0.0030878 mol / (11.10 mL + 70.90 mL) [converted to liters]
= 0.0030878 mol / 0.082 L
= 0.0377 M
Now that we have the concentration of NH4+, let's calculate the pOH using the pKa value for NH4:
pOH = pKa + log(concentration of NH4+ / concentration of NH3)
= 9.241 + log(0.0377 M / 0.1300 M) [concentration of NH3 is given, so we don't need to calculate it]
= 9.241 + log(0.2892)
= 9.241 + (-0.537)
= 8.704
Finally, we can find the pH by subtracting the pOH from 14:
pH = 14 - pOH
= 14 - 8.704
= 5.296
So, the pH of the solution made from mixing 11.10 mL of 0.278 M HCl solution with 70.90 mL of 0.1300 M ammonia solution is approximately 5.296.