A0.500-l sample of h₂so₄ solution was analyzed by taking a 100.0ml aliquot and adding ml of . after the reaction occurred, an excess of ions remained in the solution. the excess base required ml of for neutralization. calculate the molarity of the original sample of h₂so₄. sulfuric acid has two acidic hydrogens.

The molarity of the original sample of H2SO4 is 0.009289 M

Explanation:

Step 1:  The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 2: Calculating moles of NaOH

moles of NaOH = Molarity of NaOH * Volume of NaOH

moles of NaOH = 0.213 M * 50*10^-3 L

moles of NaOH = 0.01065 moles

Step 3: Calculating moles of HCL

HCl + NaOH → NaCl + H2O

13.21 mL of 0.103 M HCl

moles of HCl = molarity of HCl * Volume of HCl

moles of HCl = 0.103M * 13.21 *10^-3 L

moles of HCl =0.00136 moles

Step 4: Calculating moles of NaOH in excess

For 1 mole of HCl consumed there is 1 mole of NaOH needed to produce 1 mole of NaCl and 1 mole of H2O

For 0.00136 moles of HCl consumed, there is also 0.00136 moles of NaOH needed. Those are 0.00136 moles NaOH in excess.

Step 5: Calculating moles of NaOH reacted

0.01065 moles - 0.00136 moles = 0.009289 moles of NaOH reacted

Step 6: Calculating moles of H2SO4

There is 2 moles of NaOH consumed per 1 mole of H2SO4 consumed.

For 0.009289 moles of NaOH reacted, there is needed 0.009289/2 moles = 0.0046445 moles of H2SO4

Step 7: Calculating  molarity of H2SO4

Molarity of H2SO4 = moles of H2SO4 / volume of H2SO4

Molarity of H2SO4 = 0.0046445 / 0.5L = 0.009289 M

The molarity of the original sample of H2SO4 is 0.009289 M

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante