Idk how to do this. This is my attempt. I\'m probably headed in the wrong direct

Question

Idk how to do this. This is my attempt. I'm probably headed in the wrong direction because I don't know how to give an equation in terms of [H3O+]. I need help on all 4 parts please!

Silvina a as ases, an solution (strong/weak acid/base or buffer) you get in each of (assuming you had a table of Ka's.) 5. Tell me what type of acid/base pesh (stro for LH.O'1 based on easily obtained values uation for ne you had a table of kat A.) Mix 20 ml of 0.1 M NaOH with 35 mL of 0.05 M HCI .20. C) 1-2mmol UaCH 0.26 mg of sodium cyanide in a flask, add 50 mL of water and shake well C) Treat 10 mmol of solid sodium carbonate with 20 mmol of aqueous hydrochloric acid, resulting in a total solution volume of 25 mL. Necohel wAur gA D) Mix 25 mL of 0.1 M aqueous ammonia with 25 mL of 0.1 M ammonium chloride.

Explanation / Answer

a)

mol of NaOH = MV = 0.1*20*10^-3 = 0.002

mol of HCl = MV = 0.05*35*10^-3 = 0.00175

mol of NaOH left =0.002- 0.00175 = 0.00025

get OH-] = mol of OH- / Vtotal = 0.00025/((20+35)*10^-3) = 0.004545 M

Kw = [OH-][H3O+]

[H3O+] = (0^-14)/(0.004545) = 2.2*10^-12 M

this is NOT a buffer, only strong base in water solution

b)

Sodium cyanide --> NaCN --> NA+ + CN-

CN- + H2O <-> HCN + OH-

this is basic, weak base, no buffer formation

Kb = [HCN][OH-]/[CN-]

Kb = Kw/Ka

Kw/Ka = [HCN][OH-]/[CN-]

[H3O+] can be obtained once [OH-] data is obtained

[H3O´+] = Kw/[OH-]

c)

Na2CO3, HCl

this is complete neutralization

CO3-2 + H2O <-> HCO3- + OH- forms

Kb = [HCO3-][OH-]/[CO3-2]

Kb = Kw/Ka2

Kw/Ka2= [HCO3-][OH-]/[CO3-2]

solve for OH- then solve for H3O+

D)

this is a buffer, (basic) since weka base, NH3 and its conjugate acid, NH4+ re present

pH = pKa + log(NH3/NH4)

[H3O+] = 10^-(pH) = 10^-(pKa + log(NH3/NH4))

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