A #25*mL# volume of sodium hydroxide solution of #0.150*mol*L^-1# concentration is stoichiometrically equivalent to a #15.0*mL# of sulfuric acid. What is #[H_2SO_4]#?

Question

A #25*mL# volume of sodium hydroxide solution of #0.150*mol*L^-1# concentration is stoichiometrically equivalent to a #15.0*mL# of sulfuric acid.  What is #[H_2SO_4]#?

Peyton Adams · Accepted Answer

To find #[H_2SO_4]#, we need (i), a stoichiometrically balanced equation:

Na_2SO_4(aq) + 2H_2O(l)# + #H_2SO_4(aq) + 2NaOH rarr. additionally (ii) comparable amounts of sodium hydroxide, #"Moles of NaOH"=27.75x10^-3*mol#, 25.0*mLx10^-3*L*mL^-1x0.150*mol*L^-1 We KNOW that the initial volume of #H_2SO_4(aq)# contained half an equiv of sulfuric acid based on the equation. Number [H_2SO_4] therefore equals (1/2xx3.75xx10^-3*mol)/(15.0xx10^-3L)~=0.13*mol*L^-1#. Here, all I've done is utilize the connection: Consequently, #"Volume"="Moles of solute"#, "Concentration"xx, etc., etc.

Mateo Aguilar · Answer

undefined To find the concentration of sulfuric acid ([H2SO4]), use the stoichiometric equivalence between sodium hydroxide (NaOH) and sulfuric acid (H2SO4):

Moles of NaOH = Volume of NaOH solution (in L) × Concentration of NaOH (in mol/L)
Moles of NaOH = 25 mL × 0.150 mol/L = 3.75 mmol

Since the stoichiometric ratio between NaOH and H2SO4 is 2:1, the moles of H2SO4 are half of the moles of NaOH.

Moles of H2SO4 = 3.75 mmol / 2 = 1.875 mmol

Now, we'll find the concentration of H2SO4:

[H2SO4] = Moles of H2SO4 / Volume of H2SO4 solution (in L)
Volume of H2SO4 solution = 15 mL = 15 × 10^-3 L

[H2SO4] = 1.875 mmol / 15 × 10^-3 L = 125 mol/L