Mastering Chem Questions 1) A beaker with 1.2010 2 mL of an acetic acid buffer w
ID: 811171 • Letter: M
Question
Mastering Chem Questions
1) A beaker with 1.20102mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 9.00mL of a 0.460M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
2)When Zn(OH)2(s) was added to 1.00 L of a basic solution, 1.1010?2mol of the solid dissolved. What is the concentration of OH? in the final solution?
3)Calculate the molar solubility in NaOH. Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.200M NaOH?
4)What is the pH change of a 0.280M solution of citric acid (pKa=4.77) if citrate is added to a concentration of 0.125M with no change in volume?
5)Imagine that you are in chemistry lab and need to make 1.00L of a solution with a pH of 2.40. You have in front of you, 100 mL of 7.0010?2M HCl,100 mL of 5.0010?2M NaOH, and plenty of distilled water. You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 84.0mL of HCl and 90.0mL of NaOH left in their original containers.Assuming the final solution will be diluted to 1.00L , how much more HCl should you add to achieve the desired pH?
Explanation / Answer
1. Using H-H equation,
5.0 = 4.76 + log ([acetate]/ [acetic acid])
1.74 [acetic acid] = [acetate]
Also, given: acetate+ acetic acid = 0.1M
So, 1.74 [acetic acid] + [acetic acid] = 0.100
[acetic acid] = 0.0365 M
[acetate]= 0.1M - 0.044M =0.0635 M
Now, no. of moles of acetic acid present = 0.0365 M x 0.12 L = 0.00438 moles
no. of moles of acetate present = 0.0635 x 0.12 L = 0.00762moles
moles of H+ added = no.of moles of HCl = molarity x volume in liter = 0.64 M x 0.009 =0.00576 moles
So, acetate ions will react with H+ to give more of acetic acid as CH3COO- + H+ = CH3COOH
new moles of acetate = 0.00762 - 0.00576 = 0.00186 moles
new moles of acetic acid = 0.00438 + 0.00576=0.01014 moles
pH = 4.760 + log (0.00186/ 0.01014) = 4.02
pH change = 5.00 - 4.02 = 0.98
pH change is the decrease of 0.98