A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 30.0 mL of base to reach the equivalence point. (a) What is the molar mass of the acid?

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Hello everyone today, we're being given the falling problem a 0.2850 g sample of an unknown mono protic acid was dissolved in 30 mL of water. The resulting solution was tied with 300.115 Mueller of potassium hydroxide solution, Calculate the molar mass of the acid. If a total of 27.5 ml of potassium hydroxide was used to reach the equivalence point. So first we want to make note of the fact that we are using a mono product acid and then a mono product acid Simply just has one hydrogen. And so this will result in the following reaction that mono product acid will result react with that potassium hydroxide. Reform that potassium with that acid as well as water. The second thing you want to know is what it means to be at an equivalence point. And so when you're at an equivalence point, that essentially just means that the moles of your acid are going to equal the molds of your face until third. To figure out the molar mass, we need to do some calculations. So we have to use our mole polarity equation, we're just saying that the polarity of our acid times the volume of our acid is going to equal the malaria of our base times the volume of our base. So to find the molds of acid, We simply need to take the middle leaders of that potassium hydroxide that we got. So we had that 27.5. Middle leaders have to convert it to leaders so we can see that one Middle leader is equal to 10 to the negative 3rd leaders. And then we can finally multiply that by the polarity which is 0.115. And this is the polarity of the potassium hydroxide. Of course moles over leaders. And so our units will cancel out and will be left with 3.1625 times 10 to the negative third moles 4th. We want to recall what our molar mass equation is. Mm That's going to be mass over our moles. Simplifying that we take our masks that we have which is our 0.2850 g that our sample is. And we divide that by the moles. That we just calculated for 3.1625 times 10 to the negative third moles. To give us a final molar mass of 90.1 g per mole. As our answer. I hope this helped. And until next time.

A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 30.0 mL of base to reach the equivalence point. (a) What is the molar mass of the acid?

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Key Concepts

Monoprotic Acid

Titration and Equivalence Point

Molar Mass Calculation