10) The plH of a solution prepared by dissolving 050 mol of solid methylamine 10

Question

10) The plH of a solution prepared by dissolving 050 mol of solid methylamine 10) CH3NH3C) in 1.00 L of 1.0 M methylamine (CH3NH2) is 1.0 x 10-4. (Assume the final volume is 100 L) A) 2.86 B) 10.28 11.14 D) 10.31 E) 16 11) What change will be caused by addition of a smail amount of HC to a solution containing fluoride 11) ions and hydrogen fluoride? A) The coneentration of duoride ions will increase as will the concentration of hydroniums ions 18) The concentration of hydrogen fluoride will decrease and the concentration of fluoride ions will increase. C) The concentration of Quoride lon will decrease and the concentration of hydrogen fAluoride will increase D) The concentration of hydroniums lons will increaue significanly E) The fluoride ions will precipitate out of solution as its acid salt 120 100 sclueon 80 20 Equivalenco Poire in flask 80 40 5 10 15 20 25 30 35 40 45 TL of 0.115 M NaOH added to flask 12) A 25.0 mL sample of a sulution of a monoprotic acid is titrated with a 0.115 M NaOHH solution. The 12) utration curve above was obtained. The concentration of the monoprotic acid is about mol/L 0.100 D) 0.120 E) 25.0 A) 0.240 H) 0.0600

Explanation / Answer

Q10

first, get the concentration

[CH3NH4Cl] = mol/V = 0.5/11 = 0.5 mol of CH3NH4Cl

[CH3NH3] = 1 M

then,

this is a buffer

pOH = pKb + log(CH3NH4Cl/CH3NH3)

pOH = -log(10^-4) + log(0.5/1)

pOH = 3.69

pH = 14-3.69 = 10.31

then, from the list, choose D

Q11

if we add HCl to : F- ions in HF

this is a buffer, the pH will increase slightly due to the F- + H+ = HF formation equilbrium

choose C

Q12

mmol of base = MV = 0.115*22= = 2.53

mmol of acid = mmol of base = 2.53

[Acid] = mmol/mL = 2.53/25 = 0.10M

choose C, since it is the nearest

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