chemistry question need help?

Question

What volume (mL) of 0.204 M potassium hydroxide solution would just neutralize 32.8 ml of 0.123 M H2SO4 solution?

Answer 1

The easiest way to do neutralization calculations is to use M1V1n1 = M2V2n2 where M = molarity; v = volume, and n = the number of hydrogen ions that the acid can donate or the number of hydroxide ions in your base. For your question, (V1 * .204 * 1) = (32.8 * .123 * 2). Divide both sides of the equation by (.204 *1) to isolate your base volume.

Answer 2

1. Calculate moles of H2SO4 from the volume (in Liters) and the molarity.

2. Recognize that 2 moles of KOH are required to react with 1 mole of H2SO4

3. From the moles of KOH required and its concentration, calculate the volume of KOH.

4. Convert volume of KOH into mL

Answer 3

potassium hydroxide = KOH

acid + base = water + salt

KOH + H2SO4 --> HOH + 2 K2SO4

molarity = moles of solute/volume of solution (in litres)
therefore you have .123(.0328) = .0040344 moles of H2SO4

you need that many moles x 2 of KOH to neutralise

volume of solution = moles of solute x molarity
2(.0040344)(.204) = .0016430352 litres
= 1.6430352 mL

i didn't inclde significant digs, but you should be able to figure that out yourself