Copy of a buffer consists of 0.33 m na2hpo4 and 0.28 m na3po4. given that the k values for h3po4 are, ka1 = 7.2 x 10-3, ka2 = 6.3 x 10-8, and ka3 = 4.2 x 10-13, calculate the ph for this buffer

Use Ka3

henderson-hasselbach equation: pH = pKa + log [base]/[acid]

pKa3 = - log Ka3 = - log 4.2 x 10^-13 = 12.38

therefore: pH = 12.38 + log (0.28/0.33) = 12.30

the pH is 12.30

Explanation:

phosphoric acid is a polyprotic acid meaning it donates more than one proton

weak Acid  ↔ conjugate Base

H3PO4      ↔   H2PO4^-  corresponding to Ka1

H2PO4^-     ↔   HPO4^2- corresponding to Ka2

HPO4^2-    ↔   PO4^3-   corresponding to Ka3

A buffer consist of a weak acid and its conjugate base, the given buffer has the combination of HPO4^2- and PO4^3- thud we used Ka3

knowing that we used henderson-hasselbach equation to get the pH which is 12.30

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

um i think it might be a car?