Question

if 10.83 mL of an acetic Acid required 9.34 mL of 1.0185 M NaOH to reach the endpoint, what’s the concentration of Acetic Acid solution?

Answer 1

Solution: - . Since, we have been not provided the ka values of Acetic Acid, hence, assuming 100% dissociation, we have Volume of acetic Acid - 10 % 3 ml = ? Concentration a acetic Acid.

S of Acid Volume of NaOH required 9.34 ml concentration of NaOH = 1.0185m Now, We know, - Movis M2 V2 A for Acid . for base : molanity= M,=? via Volume = 10.83 me of Acid mz> molarity Cone.) n Naoul = 1.0185m V2 > Volume of Naon. = 9-34 ml O Required Substituting all values, We get, mi X 10.83 5 1.0185x9:34 06. Mis 9.513 = 0.878 M DE " 10.83 07, M, = 8.78x109 m. 60.67839 m) 8. Cunce of acetic And solution is 0.8784M
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