At equivalence there is no more HA and no more NaOH, for this particular reaction. So that means we have a beaker of NaA and H2O. The H2O contributes 1 x 10-7 M hydrogen ion and hydroxide ion. But NaA is completely soluble because group 1 ion compounds are always soluble. So NaA breaks apart in water and it just so happens to be in water. So now NaA is broken up. The Na+ doesn't change the pH but the A- does change the pH. Remember that the A anion is from a weak acid. That means it will easily attract a hydrogen ion if one is available. What do you know? The A anion is in a beaker of H+ ions! So the A- will attract H+ and become HA. When this happens, it leaves OH-, creating a basic solution, as shown below.
As we know methane is an Alkane to separate an Alkane We use fractional distillation to put each liquid into its fraction with the right temperature group
