Can you help me answer questions 19-20 zoom in read the question QUESTION 18 Wha
ID: 1035179 • Letter: C
Question
Can you help me answer questions 19-20 zoom in read the question QUESTION 18 What is the pri of a bulter prepared fram 033 M HCN and 109 M CN49 10 10 Write your answer using 2 places to the right of the decimal QUESTION 19 The equlibihum pi of an aquecus solution of weak acd is 3.15. if the invtial concentration of the weuk acid is 0.800 M, what is the Ka of the ac Sax 10 12x 103 20 x 106 63x 10 QUESTION 20 A student titrates 25.00 m of 0 3421 M NaCH with 0 1768 MHCL What is the pH of thtit atsion solution after the adtition of 25.00 m of rant T-25Explanation / Answer
18.
Ka = 4.9 x 10-10
So, pKa = - log Ka
= - log (4.9 x 10-10)
= 9.31
Using Henderson-Hesselbalach equation
pH = pKa + log { [salt] / [acid] }
= pKa + log { [NaCN] / [HCN] }
= 9.31 + log (0.109 / 0.833)
= 9.31 + log (0.130852341)
= 9.31 + (-0.88)
= 8.43
19.
Let the weak acid be HA
HA = H+ + A-
IC: 0.800 0 0
C: - x + x + x
EC: 0.800 – x x x
So, Ka = [H+] [A-]/ [HA]
Now,
pH = 3.15
or, - log[H+] = 3.15
or, log[H+] = -3.15
[H+] = 10-3.15
[H+] = 0.00071
So,
[A-] = [H+] = 0.00071
[HA] = 0.800 – 0.00071 = 0.79929
So,
Ka = [H+] [A-]/ [HA]
= (0.00071) x (0.00071) / (0.79929)
= 6.3 x 10-7
20.
The reaction equation for HCl and NaOH is expressed as
HCl + NaOH -----> NaCl + H2O
In the above reaction; HCl and NaOH reacts in 1:1 ratio.
Now,
25.00 mL of 0.3421 M NaOH
So, moles of NaOH = 0.3421 M x 0.025 L = 0.0085525 moles
25.00 mL of 0.1768 M HCl
So, moles of HCl = 0.1768 M x 0.025 L = 0.00442 moles
So, 0.00442 moles of HCl will neutralize 0.00442 moles of NaOH.
Hence, unreacted moles of NaOH = 0.0085525 moles – 0.00442 moles
= 0.0041325 moles
Total volume = 25 mL + 25 mL = 50 mL = 0.050 L
So, [NaOH] = 0.0041325 moles / 0.050 L = 0.08265 M = [OH-]
pOH = - log [OH-]
= - log (0.08265)
= 1.08
pH = 14 – pOH
= 14 – 1.08
= 12.92