Question

7. + -19 points My Notes A 2.256-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 35.0 mL of this solution was titrated with 0.08152-M NaOH. The pH after the addition of 25.80 mL of base was 6.74, and the equivalence point was reached with the addition of 48.31 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What is the molar mass of the acid? g/mol c) What is the pka of the acid? pka =

Answer 1

Molony o 35 0mL oj acicl Salchon e M Hanca at equivalun ca point mmalas oa Baso mmales ot Acid (MV)Acid CMU) Bayse )Mx 35.0 mL 4 31 m) (0. og)52 m) O. \ 2521 M Miimales ot acic in tn onginal Salicl scumplo (o.1 2521 M) (ı oo o mi) 11. 2521 mmals a) b) malor mass oj oejd = Ma 2.256 g 1.25 21 X to3 mals - 2.256 11-2521 X 1o3 a/may 1 25.80ml of o.o8152 m NaoH contai ng 2.l03216 mmal HA Contain= 3.938225 mmle O 112521 M 35.0 mL o A° + H2o Cg) HA NaoH 2l03216 duplkal 3.38235 2.103216 thal 1-835019 Pag pH HA 2.l03216 1-835019 PRa o 059 2 4 G4 phad vits. 68 CS amScanner