Phosphorous acid, h 3 po 3 ( aq ) , is a diprotic oxyacid that is an important compound in industry and agriculture. p k a1 p k a2 1.30 6.70 calculate the ph for each of the points in the titration of 50.0 ml of 1.8 m h 3 po 3 ( aq ) with 1.8 m koh ( aq ) .

0.0052. 1.3. 4.0. 6.7. The second EQ point

Explanation:

We can use the and   to estimate the values of   and . Therefore:

= = 0.0501

= = 1.995*10^(-7)

The value of is largely more than thus, the effects of second ionization and   can be ignored. Therefore:

⇒

I     1.8                                     0                        0

C    -x                                      x                         x

E    1.8-x                                   x                         x

=

1.3 = x*x/(1.8-x)

x^2 + 1.3x - 2.34 = 0

using x = [±b - sqrt(b^2 -4ac)]/2a

a = 1, b = 1.3, and c = -2.34

x =  [-1.3 ± sqrt(1.69 + 9.36)]/2 = (-1.3 ± 3.324)/2

Since x can only be positive, we have:

x = 1.012 M

So pH = 0.0052

When 25 ml KOH is added, the moles added is half the moles of , therefore, pH is equivalent to pKa1 which is equal to 1.3

when 50 ml KOH is added, pH = (1.3 + 6.7)/2 = 4.0

when 75 ml KOH is added, pH is equivalent to pKa2 which is equal to 6.7

when 100 ml KOH is added, the number moles of is two times the moles of and this corresponds to the second EQ point.

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation:

the answer is d : ) have a good da