How many ml of 0.100 m naoh are needed to neutralize 50.00 ml of a 0.150 m solution of ch3co2h, a monoprotic acid?

50ml = 50cm³ = 0,05dm³
C = n/V
n = 0,15 mol/dm³ × 0,05dm³
n = 0,0075 mol CH₃COOH

CH₃COOH + NaOH > CH₃COONa + H₂O
1 mol1 mol

1 mol CH₃COOH 1 mol NaOH
0,0075 mol CH₃COOH --- X
X = 0,0075 mol NaOH

V = n/C
V = 0,0075/0,1
V = 0,075 dm³ = 75cm³ = 75 mL

The answer is c. 320 g

Answerhund's rule;