Quiz #4

 

1.               The total concentration of ions in a 0.250 M solution of HCl is 0.500 M.

HCl H+(aq) + Cl-(aq);  the concentration of H+ is 0.250 M and the concentration of Cl- is 0.250 M, the sum is 0.500 M.

 

2.               A strong electrolyte is one that ionizes completely in solution.

 

3.               What is the concentration (M) of sodium ions in 4.57 L of a .398 M Na3P solution?

Na3P 3Na+ + P-3; the stoichiometric ratio of Na+         ions to Na3P is 3:1.  So the concentration of Na+ ions is 3 times the Na3P concentration, 3 x 0.398 M = 1.194 M.

 

4.                What are the spectator ions in the reaction between KOH (aq) and HNO3 (aq)?

KOH (aq) + HNO3 (aq) K+ (aq) + OH- (aq) + H+ (aq) + (NO3)- (aq) K+ (aq) + HOH or H2O (l) + (NO3)- (aq).  The only ions which do not react are K+ and NO3-, answer C.

 

5.               The compound NH4Cl is a weak acid.

True, NH4+ can donor a proton and is therefore an acid and since it isnít one of the memorized strong acids it must be a weak acid.

 

6.               Rusting of iron is an oxidation reaction.

 

7.               Ca(OH)2 is a strong base?

True, memorized strong base.

 

8.               A 36.3 mL aliquot of 0.0529 M H2SO4 (aq) is to be titrated with 0.0411 M NaOH (aq).  What volume (mL) of base will it take to reach the equivalence point?

 

The equivalence point is reached when the moles of acid equals the moles of base.  (36.3 mL) x (1 L/1000 mL) x (0.0529 M H2SO4) = 0.00192 moles of H2SO4.  Since H2SO4 is a diprotic acid we need twice as much NaOH.  So we need 0.00192 x 2 = 0.00384 moles of NaOH.  (X mL NaOH) x (1 L/1000 mL) x (0.0411 M NaOH) = 0.00384 moles.  So, X mL NaOH = (1000 mL/1 L) x (0.00384 moles/0.0411 M NaOH) = 93.6 (mL).

 

9.               What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to 200.0 mL to make a 1.50 M solution of sodium hydroxide?

 

(x mL NaOH (conc.)) x (1 L/1000 mL) x (6.00 M NaOH) = (200.0 mL NaOH (dil.)) x (1 L/1000 mL) x (1.50 M NaOH);   x mL NaOH (conc.) = (200.0 mL NaOH (dil.)) x  (1.50 M NaOH/6.00 M NaOH); x mL NaOH (conc.) = 50.0 (mL).

 

10.         Oxidation is the loss of electrons and reduction is the gain of electrons.