Can you please solve questions 2 and 3? Thank you! A Back-titrate with additiona
ID: 578061 • Letter: C
Question
Can you please solve questions 2 and 3?Thank you! A Back-titrate with additional NaOH to neutralize the additional HCI that was added. (B) Back-titrate with additional HCI to neutralize the additional NaOH that was added (C) Start over again, being careful to stop the titration just before the equivalence point. (D) Perform an additional two trials and average the volumes of the NaOH for all three trials. 2. A student uses acetic acid to titrate a solution of copper (1) hydroxide to determine the concentration of the copper (I) hydroxide. If 30.0 mL. of 0.250 M acetic acid is needed to neutralize 15.0 ml copper (1) hydroxide, what is the molarity of the copper (1) hydroxide? (A) 0.125 M B) 0.250 M HA is a weak, monoprotic acid that dissociated according to the equation (C) 0.375 NM (D) 0.500 M 3. HA(aq) H+(aq) + A-(aq) and Ka 4.0 x 10" If the initial concentration of HA is 0. 10 M, find the concentration of H'. I U.tU (A) 2.0 x 104M (B) 2.0 x 10-5 M (C) 7.5 x 108 M HA 8.5 x 10 10M Questions 4 and 5 refer to the ollowing reaction: CH3NH2(aq) + H20() CHsNH3+(aq) + OH-(aq) Kb = 4.00 x 10" 4. Determine the [OH] for 0.25 M methylamine, CH:NH2. (A) 2.0 x 10-7 M (B) 6.0 x 105 M (C) 1.0 x 104 M (D) 1.0 x 102 M 5. Determine the pH of the methylamine solution.
Explanation / Answer
2) CuOH(aq) + CH3COOH(aq) ----> CH3COOCu(aq) + H2O(l)
1 mol CuOH(aq) = 1 mol CH3COOH(aq)
no of mol of CH3COOH = 30*0.25 = 7.5 mmol
NO of mol of CuOH = 7.5 mmol
molarity of CuOH = 7.5/15 = 0.5 M
answer: D.0.5 M
3) we know that,
Ka = cx^2
(4*10^-9) = 0.1*x^2
x = degree of dissociation = 0.0002
[H+] = CX
= 0.1*0.0002
= 2*10^-4 M
answer: A.2*10^-4 M