Question

What is the concentration of Al(OH)3 when 300 mL of it will neutralize 250. mL of 4.0M HNO3?

If someone could walk me through this, that would be great

Answer 1

First make a balanced equation for your problem.

Answer 2

I'm kinda learning this right now but this is how I'll approach this problem

Al(OH)3 + 3HNO3 -> Al(NO3)3 + 3H2O

M = moles / volume (L)
Convert your mL solutions to L in order to plug them into formula.

250. mL * (1 L/ 1000 mL) = .25 L

.25 L *. 4 M = 1 mol of HNO3

HNO3 then multiply times mole ratio to get moles of Al(OH)3
.
1 mol of HNO3 * 1 mol Al(OH)3 / 3HNO3 = .333 mol Al(OH)3


300 mL -> .3L

M = .333 mol Al(OH)3 / .3 = 1.11 M Al(OH)3

Someone might wanna double check this answer. I wanna make sure I got it right.