Astudent dissolves 0.2975 g of the unknown weak acid to give 100. ml solution. s/he obtains the following titration curve for his/her best titration. at the equivalence point, 19.0 ml of 0.1000 m naoh was added.(a) calculate the molar mass of the acid. g/mol(b) what is the pka of the acid?

MM = 156.57 g/mol

pH at V = 9.5 mL will be the pKa value for the acid

Explanation:

solution

for the molar mass  MM is here express as

MM =   .....................1

here  

mass = 0.2975  g

and

mol = MV = 0.1000 × 19 ×

MV = 0.0019 mol of acid

so

MM =  

MM = 156.57 g/mol

and

we know for pKa the easiest way is to find the half equivalence point  is

V half = 0.5 × Vtitration

V half =  0.5 × 19

V half  = 9.5 mL

so  pH at V = 9.5 mL will be the pKa value for the acid

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

i think that the answer is c