Download 10 Solved Questions on Concentration of Acids and Solutions | CHEM 1110 and more Exams Chemistry in PDF only on Docsity!
Solutions Exercises
Purpose: The purpose of this exercise is to give the student practice in problems requiring calculations of solution concentrations.
Do not hand in this work sheet. When you are ready, you will be given an examination over this material. Complete the examination by your self and hand it in to receive credit.
- Calculate the concentration of a solution made by diluting a 0.125 M solution of HCl from 25 mL to 250 mL. SET-UP:
Using CV = n parametrically, C 1 V 1 = C 2 V 2
ANS: 0.0125 M
- Calculate the concentration of an acetic acid solution prepared by mixing 13.5 mL of 10.0 M acetic acid with 250.0 mL of 0.15 M NaCl solution. SET-UP:
Final volume is 263.5 mL
ANS: 0.51 M
- For problem 2, what is the resultant NaCl concentration? SET-UP:
ANS: 0.14 M
- Calculate the concentration of each ion in a solution prepared by mixing 200.0 mL of 0.20 M NaCl with 50.0 mL of 0.50 M KNO 3. SET-UP:
[NaCl] = 0.16 M ∴ [Na+^ ] = [Cl−^ ] = 0.16 M
[KNO 3 ] = 0.10 M ∴ [K +] = [NO 3 −^ ] = 0.10 M
Na +^ = 0.16 M
Cl -^ = 0.16 M
K+^ = 0.10 M
NO 3 -^ = 0.10 M
For the following questions, refer to the reaction:
NaOH + HCl → NaCl + H 2 O
- If 25.0 mL of 0.1025 M NaOH is reacted completely with HCl, how many moles of NaCl are produced? SET-UP:
nNaOH = (0.1025 M)(25.0 mL) = 2.56 mmol (what doesmmol mean?)
nNaOH /1 = n (^) NaCl/1 from NaOH + HCl → NaCl + H 2 O
ANS: 2.56 x 10 −3^ mol
- If 45.0 mL of 0.1025 M NaOH is reacted completely with HCl and the resultant volume is 125 mL, what is the molarity of the resultant NaCl solution? SET-UP:
nNaOH = 4.61 mmol = n (^) NaCl
M (^) NaCl = 4.61 mmol/125 mL
ANS: 3.69 x 10 −2^ M
(Still referring to the reaction: NaOH + HCl → NaCl + H 2 O )
- A 0.1064 M NaOH solution was used to find the concentration of an HCl solution. 20.00 mL of the HCl solution was used. In order to reach an end point 18.42 mL of the NaOH solution was used. What is the concentration of the HCl? SET-UP:
Careful with methodology! This is not a dilution problem. (See question 9)
ANS: 0.1155 M
- For the reaction: 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O , how many liters of 1.05 M NaOH are required to exactly neutralize 224 L of 1.05 M H 2 SO 4? SET-UP:
Note here: nNaOH /2 = n (^) H2SO4/1!
ANS: 448 L
- Benzoic acid has a molecular weight of 122.13 g/mol and has one acidic hydrogen. In a standardization procedure, 0.443 g of benzoic acid was used in a titration with a NaOH solution. The end point was observed after using 35.40 mL of NaOH solution. What was the concentration of the NaOH? SET-UP:
let HX ≡ benzoic acid
HX + NaOH → NaX + H 2 O and nHX/1 = n (^) NaOH /
nHX = 0.443 g/122.12 g/mol = 3.63 x 10−3^ mol = nNaOH
M (^) NaOH = 3.63 x 10 −3^ mol/35.40 x 10 −3^ L =
ANS: 0.1025 M
