Question

Part A

7.6×10−3 M HBr,

Express your answer using two decimal places.

pH =

Part B

1.39 g of HNO3 in 550 mL of solution,

Express your answer using three decimal places.

pH =

Part C

2.80 mL of 0.290 M HClO4 diluted to 55.0 mL ,

Express your answer using three decimal places.

pH =

Part D

A solution formed by mixing 14.0 mL of 0.110 M HBr with 20.0 mL of 0.220 M HCl.

Express your answer using three decimal places.

pH =

Part E Calculate the concentration of an aqueous solution of NaOH that has a pH of 11.57. Express your answer using two significant figures.

Answer 1

A) 7.6×10−3 M HBr,

[HBr] = [H+] since 100% ionization

pH = -log[H+]

pH = -log(7.6*10^-3)

pH = 2.12

b)

1.39 g of HNO3 in 550 mL of solution,

mol of HNO3 = mass/MW = 1.39/63.0128 = 0.0220 mol

[HNO#] = mol/V = (0.0220) / (0.55) = 0.04 M

[H+] = [HNO3] = 0.04M

pH = -log([H+])

pH = -log(0.04)

pH = 1.30

c)

2.80 mL of 0.290 M HClO4 diluted to 55.0 mL ,

M1*V1 = M2*V2

M2 = M1*V1/V2 = 0.29 * 2.8 / 55

[HClO4] = 0.0147636 M

[H+] = 0.0147636 M

pH = -log(0.0147636) = 1.83

D)

Total V = 14+20 = 34 mL

mmol of HBr = MV = 14*0.11 = 1.54

mmol of HCl = MV = 20*0.22 = 4.4

mmol H+ = HBr +HCl mmols = 1.54+4.4 = 5.94

[H+] = mmol/mL = 5.94/34 = 0.1747

pH = -log[H+]

pH = -log(0.1747)

pH = 0.76

E)

pH = 11.57

pOH = 14-pH = 14-11.57

pOH = 2.43

[OH-] = 10^-pOH = 10^-2.43 = 0.003715 M

[NaOH] = [OH-] = 0.003715 M

[NaOH] = 0.0037 M