From the calculations, the pH of the buffer is 3.1.
<h3>What is the pH of the buffer solution?</h3>
The Henderson-Hasselbach equation comes in handy when we deal with the pH of a buffer solution. From that equation;
pH = pKa + log[(salt/acid]
Amount of the salt = 25/1000 * 0.50 M = 0.0125 moles
Amount of the acid = 75/1000 * 1.00 M = 0.075 moles
Total volume = ( 25 + 75)/1000 = 0.1 L
Molarity of salt = 0.0125 moles/0.1 L = 0.125 M
Molarity of the acid = 0.075 moles/0.1 L = 0.75 M
Given that the pKa of lactic acid is 3.86
pH = 3.86 + log( 0.125/0.75)
pH = 3.1
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The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)
<h3>Further explanation</h3>
13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

Then mol iodine (MW=126.9045 g/mol) :

mol ratio of Cobalt and Iodine in the compound :
Chemical,when you light a match stick, you are starting a chain reaction of the substance chemical formula, making it a chemical change