What is the concentration of an unknown sulfuric acid solution if 0.2M Sodium hydroxide is titration to neutralize the acid? The initial volume of the Sodium hydroxide is .50ml and the final volume is 30.75 ml. There is 35mL of acid.

Question

What is the concentration of an unknown sulfuric acid solution if 0.2M Sodium hydroxide is titration to neutralize the acid? The initial volume of the Sodium hydroxide is .50ml and the final volume is 30.75 ml. There is 35mL of acid.

Answer 1

See your H3PO4 problem. Same kind of problem. Slightly different equation.

H2SO4 + 2NaOH => Na2SO4 + 2H2O

Answer 2

To find the concentration of the unknown sulfuric acid solution, you can use the equation:

Molarity (Sulfuric Acid) x Volume (Sulfuric Acid) = Molarity (Sodium Hydroxide) x Volume (Sodium Hydroxide)

Let's plug in the given values:

Molarity (Sulfuric Acid) x 35 mL = 0.2 M x (30.75 mL - 0.50 mL)

Multiply both sides of the equation by the volume of sulfuric acid:

Molarity (Sulfuric Acid) = (0.2 M x (30.75 mL - 0.50 mL)) / 35 mL

Simplifying the equation:

Molarity (Sulfuric Acid) = (0.2 M x 30.25 mL) / 35 mL

Molarity (Sulfuric Acid) = 0.173 M

Therefore, the concentration of the unknown sulfuric acid solution is approximately 0.173 M.

Answer 3

To determine the concentration of the unknown sulfuric acid solution, you can use the concept of stoichiometry in a neutralization reaction.

First, let's write the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH):

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.

Next, let's find the number of moles of sodium hydroxide used in the titration. To do this, we can use the formula:

moles of solute = concentration x volume

Given that the concentration of sodium hydroxide is 0.2M and the initial volume is 0.50 mL while the final volume is 30.75 mL, we can calculate the moles of sodium hydroxide used:

moles of NaOH = 0.2 M x 30.75 mL = 6.15 mmol

Since 1 mole of H2SO4 reacts with 2 moles of NaOH, we can say that 6.15 mmol of NaOH reacts with 3.08 mmol of H2SO4.

Now, let's find the concentration in Molarity (M) of the sulfuric acid solution. We can use the formula:

concentration (M) = moles of solute / volume (L)

Given that the volume of the sulfuric acid solution is 35 mL, we need to convert it to liters:

35 mL = 35/1000 L = 0.035 L

Now we can calculate the concentration of the sulfuric acid:

concentration of H2SO4 = 3.08 mmol / 0.035 L = 88 M

Therefore, the concentration of the unknown sulfuric acid solution is 88 M.