Question

7) Which one of the following is the weakest acid? A) HF (Ka = 6.8 × 10-4)

B) HClO (Ka = 3.0 × 10-8)

C) HNO2 (Ka = 4.5 × 10-4)

D) HCN (Ka = 4.9 × 10-10)

B) 5.7 x 10–2 M D) 2.9 x 10–3 M

22) Calculate the hydrogen ion concentration in a solution of iced tea with lemon having a pH of 2.87.

A) 2.9x10–2M D) 2.9x10–3M B) 5.7 x 10–2 M E) 5.7 x 10–4 M C) 1.3 x 10–3 M

23) You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [Ka = 1.8 x 10–5]

A) 4.65 B) 4.71 C) 4.56 D) 4.84

PLEASE solve all the Q.

Answer 1

D) HCN (Ka = 4.9 × 10-10)

HCN has less Ka values. So it is weak acid

22.

PH = 2.87

-log[H^+] = 2.87

[H^+] = 10^-2.87   = 0.0013M

C) 1.3 x 10–3 M >>>>answer

23.

no of moles of CH3COOH = molarity * volume in L

                                            = 0.3*0.5

                                            = 0.15moles

no of moles of CH3COONa = molarity * volume in L

                                             = 0.2*0.5

                                             = 0.1moles

no of moles of NaOH = molarity * volume in L

                                     = 1*0.02 = 0.02moles

no of moles of CH3COOH after addition of 0.02 moles of NaOH = 0.15-0.02 = 0.13 moles

no of moles of CH3COOna after addition of 0.02 moles of NaOH = 0.1+0.02 = 0.12moles

PKa = -logKa

         = -log1.8*10^-5

         = 4.74

PH   = Pka + log[CH3COONa]/[CH3COOH]

        = 4.74 + log0.12/0.13

        = 4.75 -0.03476

         = 4.71

B) 4.71 >>>>answer