Question

What volume of a 0.382 M hydrobromic acid solution is required to neutralize 19.3 mL of a 0.127 M potassium hydroxide solution?

mL hydrobromic acid

Answer 1

Given

molarity of KOH = 0.127 M = 0.127 mol/L

Volume of KOH = 19.3 ml = 0.0193 L

No. Of moles of KOH = volume * molarity = 0.0193 L * 0.127 mol/L = 2.45 * 10^-3 moles

1 Mole of KOH will give 1 mole of OH-

2.45 * 10^-3 moles of KOH will give 2.45 * 10^-3 moles of OH-

So we need 2.45 * 10^-3 moles of H+ to nuetralize 2.45 * 10^-3 moles of OH-

1 mole of HBr will give 1 mole of H+

So we need 2.45 * 10^-3 moles of HBr

no. Moles of HBr = volume * molarity

2.45 * 10^-3 mol = volume * 0.382 mol/L

Volume = 6.41 * 10^-3 L of HBr

So we need 6.4 ml of HBR

So