Answer to Question #90328 in General Chemistry for Krysha Soukoroff

Question #90328
What volume of 12.7 mol/L H2SO4(aq) solution is required to neutralize 5.00 L of 0.100 mol/L NaOH(aq) solution?
1
Expert's answer
2019-05-28T03:17:28-0400

Given:

"V_{NaOH} = 5.00\\ L""M_{H_2 SO_4} = 12.7 \\ mol\/L""M_{NaOH}= 0.100 \\ mol\/L"

Solution:



"H_2 SO_4 + 2NaOH = Na_2 SO_4 + 2H_2 O"

Clearly there is a 1:2 equivalence, so


"n_{H_2 SO_4} = \\frac{1}{2}* n_{NaOH}"

and as a first step we calculate the number of moles of sulfuric acid:


"n_{NaOH} = M_{NaOH} * V_{NaOH}""n_{NaOH} = 0.100 mol\/L * 5L = 0.5mol"

from first formula:


"n_{H_2 SO_4} = 0.5* n_{NaOH} = 0.5* 0.5 mol = 0.25 mol"

From amount of substance we can get volume:


"V_{H_2SO_4} = \\frac{n_{H_2SO_4}}{M_{H_2SO_4}} = \\frac{0.25 mol}{12.7 mol\/L} \\approx 0.2L"

Answer:

"V_{H_2SO_4} \\approx 0.2 L"


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