What is the molar mass of a 0.350 g sample of a HA acid if it requires 50.0 mL of 0.440 M Sr(OH) 2 to completely neutralize it? A is used as a place holder for the unknown nonmetal of the acid.

What is the molar mass of a point 350 gram sample of a acid if it requires 50.0 milliliters of point 440 molar Strongium Hydroxide to completely neutralize it. A is used as a placeholder for the unknown nonmetal of the acid. Alright. So first of all, we have an acid base titration here, so we need a balanced equation. And that's the only way we're gonna be able to solve this stoichiometrically. So we have our acid here. We're reacting with strontium hydroxide. Now this acid will be composed of h plus 1 and a minus 1. Strontium is s r 2+, and o h minus is minus 1. This h+minusoh would give me water plus here the numbers are different so they wouldn't cancel out, they criss cross, 1 comes here, 2 comes here, so this would be s r a 2. We need to balance this. There's 2 hydrogens here and 2 a's here, so we'd have to put a 2 here. But now I have 2 hydrogens here and 2 more here, so that's 4. So then finally I put a 2 here. So this is my balanced equation. Now, you want the molar mass of this acid, remember molar mass equals grams per mole. And technically we already have half of this figured out because we know the grams of the acid is 0.350 grams. So all we have to do at this point is figure out the moles of the unknown acid. Here, they're telling us of between the volume and molarity of strontium hydroxide. Remember, of means multiply. And we're gonna say here, remember, moles equals liters times molarity. So here I'm gonna do 0.050 liters. Molarity means moles over liters, so that's 0.440 moles of strontium hydroxide over here 1 liter. We're gonna say then, if I know the moles of strontium hydroxide, I know the moles of my acid because it's a mole to mole comparison. So moles of strontium hydroxide go here on the bottom, moles of my acid go here on top. For every 2 moles of my acid, I have 1 mole of my strontium hydroxide, So that's gonna equal 0.0 44 moles. Take that and plug it in over here. So when I do that, I'm gonna get as my molar mass 7.95 grams per mole. So that would be my final answer.

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10. Acids and Bases

Strong Acid Strong Base Titrations (Simplified)

GOB Chemistry

10. Acids and Bases

Strong Acid Strong Base Titrations (Simplified)

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Multiple Choice

What is the molar mass of a 0.350 g sample of a HA acid if it requires 50.0 mL of 0.440 M Sr(OH)₂ to completely neutralize it? A is used as a place holder for the unknown nonmetal of the acid.

15.9 g/mol

31.8 g/mol

5.30 g/mol

7.95 g/mol

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Verified step by step guidance

Start by writing the balanced chemical equation for the neutralization reaction between the acid HA and the base Sr(OH)2. The reaction is: 2 HA + Sr(OH)_2 → 2 H_2O + SrA_2.

Calculate the moles of Sr(OH)2 used in the reaction. Use the formula: moles = molarity × volume (in liters). Convert 50.0 mL to liters by dividing by 1000.

Determine the moles of HA that reacted. From the balanced equation, 1 mole of Sr(OH)2 reacts with 2 moles of HA. Therefore, multiply the moles of Sr(OH)2 by 2 to find the moles of HA.

Calculate the molar mass of HA using the formula: molar mass = mass / moles. Use the given mass of the HA sample (0.350 g) and the moles of HA calculated in the previous step.

Compare the calculated molar mass with the provided options to identify the correct answer.