Abuffer prepared by dissolving oxalic acid dihydrate ( h2c2o4⋅2h2o ) and disodium oxalate ( na2c2o4 ) in 1.00 l of water has a ph of 4.619 . how many grams of oxalic acid dihydrate ( mw=126.07 g/mol ) and disodium oxalate ( mw=133.99 g/mol ) were required to prepare this buffer if the total oxalate concentration is 0.150 m? oxalic acid has pka values of 1.250 ( pka1 ) and 4.266 ( pka2 ).

There is needed 5.8 g   of NaHC2O4 and 13.89 g of Na2C2O4

Explanation:

Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

Step 1: The equation

H2C2O4 + NaOH ⇔ NaHC2O4 + OH-     pKa1 = 1.250

NaHC2O4 + NaOH ⇔ Na2C2O4 + OH-   pKa2 = 4.266

Step 2: The pH equation

pH = pKa + log [conjugate base]/[acid]

pH = pKa + log [Na2C2O4]/[NaHC2O4]

4.619 = 4.266 + log [Na2C2O4]/[NaHC2O4]

0.353 = log [Na2C2O4]/[NaHC2O4]

10^0353 = [Na2C2O4]/[NaHC2O4]

2.25424 = [Na2C2O4]/[NaHC2O4]

[Na2C2O4] = 2.25424 *[NaHC2O4]

Step 3: Calculate the concentrations

Given is that [Na2C2O4] + [NaHC2O4] = 0.150M

So 2.25424 *[NaHC2O4] + [NaHC2O4] = 0.150M

[NaHC2O4] = 0.046 M

[Na2C2O4] = 2.25424 *[NaHC2O4] = 2.25424 *0.046 M = 0.1037 M

Step 4:  Calculate mass

0.046 moles NaHC2O4 * 126.07g/mole = 5.8 g  

0.1037 moles Na2C2O4 x 133.99g/mole = 13.89 g

k

explanation:

k

b) mutualism

hope this : )