Concentration of acid in the solution *Chemistry Help*

Need some help show me how you did it though please...

1.) A monoprotic strong acid solution was titrated with a 1.5 M sodium hydroxide solution. If 41.6 mL of the standard was needed to neutralize 50 mL of the acid solution, what was the concentration of acid in the solution?

2. ) A diprotic strong acid solution was titrated with a 0.85 M sodium hydroxide solution. If 34.6 mL of the standard was needed to neutralize 25 mL of the acid solution, what was the concentration of the acid in the solution?

more? lol

1. HA + NaOH --> H2O + NaA
1.5M NaOH x 0.0416L = 0.062moles NaOH present, 1:1 ratio with HA so 0.062moles HA neutralized
0.062moles HA / 0.05L = 1.25M HA

2. H2A + 2NaOH --> Na2A + 2H2O
0.0346L x 0.85M = 0.029moles NaOH present, 2:1 ratio
0.029moles NaOH will neutralize 0.0147moles H2A
0.0147moles / 0.025L = 0.588M H2A

Each mole of NaOH has one mole of hydroxide (OH) which nuetralizes a mole of hyrdrogen ions (H) from the monoprotic strong acid. I use conversion factors for this.

(41.6 mL NaOH solution)x(1L/1000mL)x(1.5 mol OH/1 L NaOH solution)x(1 mol H/1 mol OH)x(1 mol monoprotic strong acid/1 mol H)= 0.0624 mol strong acid

there are 0.0624 moles of monoprotic strong acid in the 50 mL volume. The concentration is 0.0624/0.05 L=1.24 M

On your second problem you use (1 mol diprotic acid/2 mol H) as a conversion factor, but you can tell your teacher that doesn't actually work because there's no such thing as a diprotic strong acid.

Just use the formula:

Molarity = concentration (moles) x volume (dm^3)

___/m\___
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