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Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Answer:
Laboratories use both distilled water and deionized water as controls in experiments. Deionization removes only non-charged organic matter from the water.
Explanation:
Distilled water removes even more impurities than deionization does, if the water undergoes a filtering process before boiling and distillation.
The reason for adding a limited amount and then an excess amount is that initially a metal hydroxide may form which becomes soluble when more base is added and the metal complex forms.
In qualitative analysis is a common to add the base in drops and then in excess. When added in drops, the metal hydroxide is formed. This metal hydroxide is often insoluble.
After this metal hydroxide is formed, the base could be added in excess such that the metal hydroxide dissolves in the excess base by forming a complex.
For instance;
CuCl2(aq) + 2NaOH(aq) -------> Cu(OH)2(s) + 2NaCl(aq)
Cu(OH)2(s) + 2OH^-(aq) -------> [Cu(OH)4]^2+(aq)
Learn more: brainly.com/question/1527403
