Oxalic acid reacts with sodium hydroxide according to the following equation:
h2c2o4 + 2 naoh → na2c20. + 2 h20
a 25.00 ml sample of the h2c2o solution required 19.62 ml of 0.341 m naoh for
neutralization. calculate the molarity of the acid.

Concentration = 0.14 M

Explanation:

Given data:

Volume of base = 19.62 mL

Molarity of base = 0.341 M

Volume of acid = 25.00 mL

Molarity of acids = ?

Solution:

Number of moles of base = Molarity × volume in litter

Number of moles = 0.341 mol/L × 0.02 L

Number of moles =  0.007 mol

Concentration of acid:

Concentration = 1/2 (0.007 mol) / 25 mL × 10⁻³ L/mL

Concentration = 0.0035 mol / 25 × 10⁻³ L

Concentration = 1.4  × 10⁻¹ M

Concentration = 0.14 M

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.