i.)Determine the pH of a solution after 20.00 mL of 0.4963 M HI has been titrate
ID: 1036259 • Letter: I
Question
i.)Determine the pH of a solution after 20.00 mL of 0.4963 M HI has been titrated with 12.64 mL of 0.5174 M NaOH.
ii.)Determine the pH of the solution when 20.00 mL of 0.1563 M aniline hydrochloride (C6H5NH3+Cl-) is titrated with 15.00 mL of 0.1249 M NaOH. (Ka(aniline)=2.40 x 10-5)
iii)Consider the titration of 25.00 mL of 0.4592 M HCl which requires 22.74 mL of KOH. Determine the concentration of KOH.
4.)Consider a solution of CHCOOH in water that has a pH of 3.56 and fill in the blanks with either increase or decrease.
If the solvent was instead a solution of NaCHOO the concentration of H+ would ____________ and the pH would _______________.
Explanation / Answer
1.)
HI is a strong acid .. So thats why it dissociates completely,
20.00 x 0.4963 = 9.926 mmole Acid
12.64 x 0.5174 = 6.540 mmole Base
subtract for excess acid
3.386 mmole H+
[H+] = 3.386 / (20.0 + 12.64) It is molar conc of H+
so, pH = - log [H+]
pH = 0.98
2.)
20 mL x 0.1563M = 3.126 mmoles aniline-HCl
0.1249M NaOH
Ka aniline-HCl = 2.4x10^-5
Kb aniline = 4.17x10^-10
C6H5NH3+ --> C6H5NH2 + H+
Kb = [C6H5NH2][H+] / [C6H5NH3+]
4.17x10^-10 = x^2 / 0.1 - x.....ignore x in denominator with Kb << 1
4.17x10^-10 = x^2 / 0.1
4.17x10^-11 = x^2
x = [H+] = 6.45x10^-6M
moles H+ = 6.45x10^-6M x 20mL = 1.29x10^-7moles H+
0.015L x 0.1249M NaOH = 0.00187moles OH-
all H+ will be neutralized, leaving 0.00187 - 6.45x10^-6 = 1.86x10^-3moles OH-
1.86x10^-3moles / (20 + 15)mL = 0.0531M
pOH = 5.32
pH = 14 - pOH = 8.672
4.)
If the solvent was instead a solution of NaCHOO the concentration of H+ would decrease and the pH would increase.
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