When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :
when Pka = - ㏒ Ka
6.86 = -㏒ Ka
∴Ka = 1.38 x 10^-7
by using ICE table:
H2PO4- → H+ + HPO4
initial 0.4 m 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka = [H+][HPO4] / [H2PO4-]
by substitution:
1.38 X 10^-7 = X^2 / (0.4-X) by solving for X
∴X = 2.3x 10^-4
∴[H+] = X = 2.3 x 10^-4
∴PH = -㏒[H+]
= -㏒ (2.3 x 10^-4)
∴PH = 3.6
Answer:
The acid must be a concentrated acid
Explanation:
Ethene is prepared in the laboratory by heating ethanol with excess concentrated tetraoxosulphate VI acid at 170°C . The reactionoccursc in two stages;
1) when the ethanol and sulphuric acid are mixed in a ratio of 1:2, ethyl hydrogentetraoxosulpate VI is formed
2) The compound formed in the first step is heated in the presence of excess concentrated sulphuric acid to give ethene and sulphuric acid.
The overall reaction can be perceived as the dehydration of ethanol. The gas produced (ethene) is usually passed through sodium hydroxide solution to remove any gaseous impurities present.
concentrated sulphuric acid is used in this process since it is a good dehydrating agent.
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Its A) Shared equally
Non-polar bonds are a type of chemical bond where two atoms share a pair of electrons with each other.
