Answer to Question #238659 in Physical Chemistry for Lubi

Question

35.6 ± 0.4 mL of 0.600 ± 0.02 M HCl was required to neutralize 35 ± 0.3 mL of a solution of NaOH.

Calculate the error in the concentration of NaOH in M from the combined errors in the data.

Answer 1

Concentration of NaOH = Volume of HCl × Concentration of HCl/Volume of NaOH

Molar mass of HCl = 36.458

0.62 ×36.458/36 = 0.63ml

0.63/36.458 = 0.0172mol/0.63 = 0.027mol/L

= 0.63 × 0.027/35.3 = 0.000482

0.62 ×36.458/35.2 = 0.64

0.64/36.458 = 0.0176/0.64 = 0.0274

= 0.64× 0.0274/34.7 = 0.00051

Error = 0.00051-0.000482 = 0.000234