Calculate the pH at the equivalence point for the titration of 0.110 M methylami

Question

Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH3NH2) with 0.110 M HCl. The K of methylamine is 5.0x 104 Number 500.0 mL of 0.110 M NaOH is added to 615 mL of 0.200 M weak acid (K resulting buffer? 5.67 x105). What is the pH of the HA(aq) + OH (aq) H2O()+A-(aq) Number Strong base is dissolved in 565 mL of 0.400 M weak acid (Ka = 4.20 x 10-5) to make a buffer with a pH of 4.19. Assume that the volume remains constant when the base is added. HA(aq) + OH (aq) H20()+A (aq) Calculate the pK value of the acid and detemine the number of moles of acid initially present Number Number pK,4.38 0.226 mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number 0.646 HA How many moles of strong base were initially added? Incorrect Look in the bottom panel for more specific feedback on how to do this part of the calculation. Number 0.4206 mol OH

Explanation / Answer

1)
Since concentration of both acid and base are same. The volume required for both to reach equivalence point will be same
Let 1 mL of both acid and base are required
Given:
M(HCl) = 0.11 M
V(HCl) = 1 mL
M(CH3NH2) = 0.11 M
V(CH3NH2) = 1 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.11 M * 1 mL = 0.11 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)
mol(CH3NH2) = 0.11 M * 1 mL = 0.11 mmol



We have:
mol(HCl) = 0.11 mmol
mol(CH3NH2) = 0.11 mmol

0.11 mmol of both will react to form CH3NH3+ and H2O
CH3NH3+ here is strong acid
CH3NH3+ formed = 0.11 mmol
Volume of Solution = 1 + 1 = 2 mL
Ka of CH3NH3+ = Kw/Kb = 1.0E-14/5.0E-4 = 2*10^-11
concentration ofCH3NH3+,c = 0.11 mmol/2 mL = 0.055 M


CH3NH3+      + H2O ----->     CH3NH2   +   H+
5.5*10^-2                    0         0
5.5*10^-2-x                  x         x


Ka = [H+][CH3NH2]/[CH3NH3+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((2*10^-11)*5.5*10^-2) = 1.049*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.049*10^-6 M



[H+] = x = 1.049*10^-6 M

use:
pH = -log [H+]
= -log (1.049*10^-6)
= 5.98
Answer: 5.98

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