Stumped on Chemistry question. 300 mLs of a 2% NaOH (sodium hydroxide, a base) ...?

2020-12-16 1:33 pm

300 mLs of a 2% NaOH (sodium hydroxide, a base) solution are mixed with 200 mLs of a 1% H2SO4 (sulfuric acid) solution.

NaOH + H2SO4 ----> H2O + Na2SO4

a) who is the limiting reagent?

b) who is the leftover reagent, and how much is left over (in moles)

c) how much products (water and salt) are made (in moles)

d) estimate the pH after the reaction is done

e) the molarity of the salt (Na2SO4) in the final solution

I am stumped by this question. How to solve it without knowing the density of each solution, and that varies by temperature?

I balanced the equation

2 NaOH + H2SO4 → 2 H2O + Na2SO4

molar masses:

NaOH: 39.997 g/mol

H2SO4: 98.07848 g/mol

H2O: 18.01528 g/mol

Na2SO4 : 142.04213856 g/mol

回答 (1)

Bobby

2020-12-16 5:03 pm

✔ 最佳答案

with dilute solution we don't need to consider density

a) 2% NaOH = 20 g / L or 20g / 40 g/ mol // L = 0.5 M NaOH
1% H2SO4 = 10g / L or 10 g/98 g /mol moles /L = 0.102 M
moles of NaOH in 300 mL = 0.5 moles/ liter * 0.300L = 0.15 moles
moles of H2SO4 in 200 mL = 0.102 moles/ liter * 0.200L = 0.020moles
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O
0.020moles if H2SO4 need 0.04 moles of NaOH but we have 0.15 moles
the NaOH is in excess and H2SO4 is the limiting reactant

b) the H2SO4 reacts with 0.04 moles of NaOH so we have 0.15 - 0.04 moles excess
= 0.11 moles of NaOH excess

c) according to our equation 0.020moles if H2SO4 gives 0.04 moles of H2O and 0.02 moles of Na2SO4

d ) after the reaction is complete we have 0.11 moles of NaOH in 500 mL of water
[OH-] = 0.11 moles / 0.5 l = 0.22 M
pOH = 0.66
pH = 13.3

e) we have 0.02 moles of Na2SO4 in 0.5L
[Na2SO4] = 0.02 moles / 0.5L =0.04 M

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