Question

A student would like to prepare a 0.010 M solution from a 1.0 M stock solution with minimal error.

Answer 1

Answer: Two-step dilution; 0.0014

Explanation:

from the question, we would be defining the parameters given.

Step 1 : consider that 10-mL transfer pipet and a 1000-mL volumetric flask is to be used.

Step 2 :  50-mL pipet and a 1000-mL volumetric flask for the first dilution, and a 100-mL pipet and a 500-mL volumetric flask is used.

given that the pipe used are 10-ml, 50-ml and 100-ml.

From the textbook on tolerance of glassware,

the tolerance at 10-ml pipet = 0.03ml

for 50-ml pipet = 0.05ml

for 100-ml pipet = 0.08ml

for 500-ml volumetric flask = 0.2ml

for 1000-ml volumetric flask = 0.3ml

from step 1, the ratio of concentration  is given as = P1/F1 = 10ml/1000ml = 0.01

while that of step 2 is given as = P1/F1 × P2/F2 = 50/1000 × 100/500 = 0.01

∴ the smallest overall relative uncertainty (Error) = σc/c

    σc/c  =  √ (Σ (σPtl/Pp)² + Σ (σFtl/Fp)²) ......................... (1)

from step 1, applying the error formula we have;

σc/c (₁)  =  √ ( (0.03/10)² + (0.3/1000)²) = 0.003015 = 3.015 × 10⁻³

also step 2, σc/c (₂) = √ (0.05/50)² + (0.08/100)² + (0.03/1000)² + (0.2/500)²

          = 0.0014 = 1.4 × 10⁻³

comparing values, we have that σc/c (₂)  ˃ ˃ σc/c (₁)  which means that the two dilution step of 0.0014 has the smallest relative uncertainty (error).

the value here is 1.4 × 10⁻³