1. Write the balanced neutralization reaction between H 2 SO 4 and KOH in aqueou

Question

1. Write the balanced neutralization reaction between H2SO4 and KOH in aqueous solution. Phases are optional.

A. 0.250 L of 0.480 M H2SO4 is mixed with 0.200 L of 0.260 M KOH. What concentration of sulfuric acid remains after neutralization?

2. The amount of I3-(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32-(aq) (thiosulfate ion). The determination is based on the net ionic equation

2S2O32-(aq) + I3- (aq) ---> S4O62-(aq) + 3I- (aq)

Given that it requires 32.8 mL of 0.470 M Na2S2O3(aq) to titrate a 15.0-mL sample of I3

Explanation / Answer

1.

H2SO4 + 2 KOH ? K2SO4 + 2 H2O

(0.250 L) x (0.400 M) = 0.100 mol H2SO4
( 0.200 L) x (0.220 M) = 0.044 mol KOH

0.044 mole of KOH would react completely with 0.044 x (1/2) = 0.022 mol H2SO4, but there is more H2SO4 present than that, so H2SO4 is in excess and KOH is the limiting reactant.

(0.100 mol H2SO4 initially) - (0.022 mol H2SO4 reacted) = 0.078 mol H2SO4 left over

Supposing the volumes to be additive:
(0.078 mol H2SO4) / (0.250 L + 0.200 L) = 0.173 M H2SO4

2.

(32.8 mL) x (0.470 M Na2S2O3) x (1 mol I3{-} / 2 mol S2O3{-}) / (15.0 mL I3{-}) =
0.231M I3{-}

3.

One mole of HCl neutralizes one mole of NH3
29 mL of 0.150M HCl contains 0.02*0.15 = 0.3 moles of HCl
The molecular mass of N is 14.00 g/mol so there were 0.0105*14.00 = 0.147 g of N in the original sample of 2.25 g. The mass percentage is then 0.147/2.25 * 100 = 6.53%

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