Question

We can measure the concentration of NaOH solution by reaction with KHP

Complete reaction with 0.9674 ± 0.0009 g of KHP (molecular weight: 204.2210 ± 0.0007 g/mol) required

27.35 ± 0.04 mL of NaOH. Find the molarity of the NaOH and its absolute uncertainty.

o rNIP Ho (NaOH) (KHP) (NaKP) Complete reaction with 0.9674 ± 0.0009 g of KHP (molecular weight: 204.2210 ± 0.0007 g/mol) required 27.35 ± 0.04 mL of NaOH. Find the molarity of the NaOH and its absolute uncertainty.

Answer 1

Given, mass of KHP = 0.9674 ± 0.0009 g, molar mass of KHP = 204.2210 ± 0.0007 g/mol

so, number of moles of KHP (n_KHP) = Mass of KHP/ Molar mass of KHP

                                                   = (0.9674 ± 0.0009) / (204.2210 ± 0.0007)

                                                   = (0.9674 ± 0.093%) / (204.2210 ± 0.000343%)   

                                                                                             [since, (0.0009 / 0.9674) *100 = 0.093%]

   = (0.9674 / 202.2210) ± (0.093% + 0.000343 %)

                                                   = 0.00474 ± 0.093343%

   = 0.00474 ± 0.0000044 moles

From the equation, it is apparent that, KHP and NaOH combine in 1:1 molar ratio,

so, n_KHP = n_{NaOH =} 0.00474 ± 0.0000044 moles

Molarity of NaOH = n_NaOH/ Volume of NaOH (mL)

                         = (0.00474 ± 0.0000044) moles / 27.35 ± 0.04 mL

                         = 1000 * (0.00474 ± 0.0000044) / 27.35 ± 0.04   mol L ^-1

                         = (4.74 ± 0.0044) / (27.35 ± 0.04) mol L ^-1

                         = (4.74 ± 0.0928%) / (27.35 ± 0.146%) mol L ^-1

                         = (4.74 / 27.35) ± (0.0928% + 0.146%) mol L ^-1

                         = 0.1733 ± 0.2388 % mol L ^-1

                         = 0.1733 ± 0.0004138 mol L ^-1

so, molarity of NaOH is 0.1733 mol L ^-1 and its absolute uncertainty is 0.0004138