Question

What is the pH of a solution obtained by adding 50.0 mL of 0.100 M NaOH (aq) to 60.0 mL of 0.100 M HCl (aq)?

Select one:

a. 11.96

b. 2.04

c. 3.11

d. 7.00

QUESTION 2

What is the pH of a solution prepared by dissolving 5.86 grams of propanoic acid, CH₃CH₂COOH (l), and 1.37 grams of NaOH (s) in enough water to make exactly 250.0 mL of solution. pK_a = 4.87 for propanoic acid?

Select one:

a. 4.58

b. 4.87

c. 4.75

d. 4.62

Answer 1

Moles of NaOH in 0.1M whose volume is 50ml =0.1*50/1000=0.005

Moles of HCl in 60ml of 0.1M= 0.1*60/10000 =0.006

Based on the reaction HCl+ NaOH---> NaCl + H2O , both ae strong bases, all the NaOH is goiing to get consumed and one is left with 0.006-0.005 moles o HCl which is 0.001 moles

Molarity =0.001*1000/(50+60)ml=0.009

pH= -log(0.009)=2.04 ( b is the correct answer)

2. Molecular weights : CH3CH2COOH = 15+14+12+32+1=74 , NaOH =40

moles of propanoic acid = Mass/ Molecular weight = 5.86/ 74 =0.077

Molarity =0.077/0.25 =0.308

Moles of NaOH= 1.37/40 =0.03425

The reaction is CH3CH2COOH+ NaOH-----> CH3CH2COONa+ H2O

1 mole of propanoic acid require 1 mole of NaOH

Propaonic acid is excess by amount =0.077-0.03425 =0.04275

Molarity =0.0427/0.25= 0.171

Moles of sodium propionate (CH3CH2COONa) formed =0.03425 molarity =0.03425/0.25 =0.137 which is sourced for CH3CH2COO- (A-)

pH= PKa+ log [A-]/[HA] = 4.87+log (0.137/0.171) =4.75 ( C is the correct answer)