When the Pka for formic acid = 3.77
and Pka = -㏒ Ka
3.77 = -㏒ Ka
∴Ka = 1.7x10^-4
when Ka = [H+][HCOO-}/[HCOOH]
when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2 by solving this equation for X
∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
= -㏒ 0.0059
= 2.23
Answer:
I believe it is "Arsenenate"
Empirical formula is the simplest way the molecular formula can be wrote so here 7 goes into all of these so it would be CH2O
Germanium is classified as a metalloid or semi-metal . (:
Answer:

Explanation:
Given:
A solution contains one or more of the following ions such as Ag,
and 
Here the Lithium bromide is added to the solution and no precipitate forms
Solution:
Since with LiBr no precipitation takes place therefore Ag+ is absent
Here on adding
to it precipitation takes place.
Precipitate is as follows,

Thus,
is present
When
is added again precipitation takes place.
Therefore the reaction is as follows,

Therefore,
are present in the solution