These were my answers and I got part marks for Q9 & Q10. I need help! I get a se

Question

These were my answers and I got part marks for Q9 & Q10. I need help! I get a second chance. Thank you!

Question 0/0.25 points You will need the molar masses of the potassium salts of phosphate /hydrogen phosphate/ dihydrogen phosphate in order to perform calculations in the lab. Determine the molar mass (in g/mol) of potassium phosphate (KPO) and enter this below Write this value, along with the molar masses of the other two salts, in the shaded cells of Table 1 in your Laboratory worksheet. Answer: 212

Explanation / Answer

8)

molar mass of K3P04 = ( 3 x 39) + ( 31 ) + ( 4 x 16) = 212

we know that

H2C03 ---> H+ + HC03- pKa1 = 6.4

HC03- ---> H+ + C032- pKa2 = 10.2

now this information should be used

9)

we know that for buffers

pH = pKa + log [conjugate base / acid ]

we can see that

pKa of the acid should be within the range of pH

given

pH = 6

so

pKa should be in the range of 6

so

the system to be used is

H2C03 ---> H+ + HC03- pKa = 6.2

so

the answer is H2C03 , HC03-

10)

similarly for pH = 10

the system to be used is

HC03- ----> H+ + C032-

the answer is HC03- , C032-

12)

now strong acid is added

we know that

acid react with the base

so

the strong acid reacts with the weak conjugate base in the buffer solution and forms weak conjugate acid

this produces very few H30+ ions in the solution and therefore only a small change in pH

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