The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the stems are used to make rhubarb pie. if pka1 = 1.23 and pka2 = 4.19, what is the ph of a 0.0321 m solution of oxalic acid?

Making the ICE table for the dissociation of oxalic acid:

H₂C₂O₄ -->  H⁺ + HC₂O₄⁻

0.0321           0        0  

 -x                 + x     +x  

0.0321-x         +x     +x  

Kₐ = x²/ 0.0321 –x  

pKₐ = - logKₐ

Kₐ = e⁻pKₐ = e⁻1.23 = 0.3

0.3 = x²/ 0.0321 – x

0.3 (0.0321 – x) = x²

-x² -0.3 x + 9.63 x 10⁻³

Using the quadratic formula:

a = -1, b = - 0.3, c = 9.63 x 10⁻³  

x = - 0.33, 0.03

Using Henderson-Hasselbalch equation

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

pH = pKₐ₁ + log [HC₂O₄⁻]/ [H₂C₂O₄]

pH = 1.23 + log (0.03/0.0321) = 1.23 - 0.03 = 1.20

Thus the pH of a 0.0321 M solution of oxalic acid is 1.20

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied