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Chapter 10 odd problems solutions
CHEM1405.P02
Dr. Stankus

Dilution of Solutions
1. What volume of (a) 12.0 M HCl is need to progress to 2.00 L of 1.00 M HCl?
Mconc x Vconc = Mdil x Vdil
(12.0 M) x (Vconc) = (1.00 M) x (2.00 L)
Vconc=0.167 L
(b) 1.04 M Na2CO3 is required to make 0.500 L of 1.00 M Na2CO3?
Mconc x Vconc = Mdil x Vdil
(1.04 M) x (Vconc) = (1.00 M) x (0.500 L)
Vconc=0.481 L

3. What volume of concentrated (18.0 M) sulphuric acid would be required to make each of the chase?
a. 1.25 L of 6.00 M solution
Mconc x Vconc = Mdil x Vdil
(18.0 M) x (Vconc) = (6.00 M) x (1.25 L)
Vconc=0.417 L

b. 575 mL of 0.100 M solution
Mconc x Vconc = Mdil x Vdil
(18.0 M) x (Vconc) = (0.100 M) x (0.575 L)
Vconc=0.00319 L

Acid-Base Titrations
5. Calculate the mebibyte of an HCl solution if 20.0 mL of it requires 33.2 mL of 0.150 M NaOH for neutralization.




7. Calculate the molarity of a Ca(OH)2 solution if 18.5 mL of it requires 28.2 mL of 0.0302 M HCl for neutralization. The products are CaCl2(aq) and piddle.





9. A 20-mL sample of gastric melted is neutralized by 25 mL of 0.10 M NaOH. What is the molarity of HCl in the fluid? Assume that all the acidity of the gastric fluid is due to HCl.



11. How many milliliters of 0.100 M H2SO4 are required to react with 10.3 mL of 0.404 M NaHCO3?




H2SO4 (aq) + 2 NaHCO3 (aq) ? Na2SO4 (aq) + 2 H2O + 2 CO2(g)

13. How many milliliters of 0.0195 M HCl are required to titrate
a. 25.00 mL of 0.0365 M KOH(aq)

b. 10.00 mL of 0.0116 M Ca(OH)2(aq),


c. 20.00 mL of 0.0225 M NH3(aq)

pH and pOH
15. What is the pH of each of the following solutions?
a. 1.0 x 10-2 M HCl
pH = -log[H3O+]= -log(1.0x10-2) = 2.00

b. 1.0 x 10-4 M HNO3
pH = -log[H3O+]= -log(1.0x10-4) = 4.00

17. What is the pOH of each of the following solutions?
a. 1.0 x 10-2 M NaOH
pOH = -log[OH-]= -log(1.0x10-2) = 2.00

b. 1.0 x 10-3 M KOH
pOH = -log[OH-]=...If you want to get a full essay, order it on our website: Ordercustompaper.com



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