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10.0 mL of 0.50 M sulfuric acid (H2SO4) neutralizes mL of 0.50 M sodium hydroxide (NaOH) solution

User Glanden
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Answer:

Step-by-step explanation:

To determine the volume of 0.50 M sodium hydroxide solution required to neutralize 10.0 mL of 0.50 M sulfuric acid, we can use the balanced chemical equation for the reaction between H2SO4 and NaOH:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

From the equation, we can see that one mole of H2SO4 reacts with two moles of NaOH.

The number of moles of H2SO4 in 10.0 mL of 0.50 M solution is:

moles of H2SO4 = (0.50 mol/L) x (10.0 mL/1000 mL) = 0.005 mol

Since the stoichiometry of the reaction is 1:2 for H2SO4 and NaOH, we need twice as many moles of NaOH to react with the H2SO4. Therefore, the number of moles of NaOH required is:

moles of NaOH = 2 x moles of H2SO4 = 2 x 0.005 mol = 0.010 mol

To calculate the volume of 0.50 M NaOH solution required, we can use the definition of molarity:

moles of NaOH = (0.50 mol/L) x (volume of NaOH/1000 mL)

Solving for the volume of NaOH, we get:

volume of NaOH = (moles of NaOH / (0.50 mol/L)) x 1000 mL = (0.010 mol / 0.50 mol/L) x 1000 mL = 20.0 mL

Therefore, 20.0 mL of 0.50 M NaOH solution is required to neutralize 10.0 mL of 0.50 M H2SO4.

User Oggy
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