The molar concentration of HCL in a solution used to remove boiler scale from pl

Question

The molar concentration of HCL in a solution used to remove boiler scale from plumbing can be determined by titration with a NaOH solution.

10 ml. of the HCL solution is placed in a 125 ml. beaker; 50 ml. of DI water are added along with 1 – 2 drops of phenolphthalein solution. It took 44.0 l. of 0.125M NaOH solution to reach an end point.

Write the reaction for the equation, calculate the number of moles of NaOH required for the titration, the number of moles of HCL in the HCL sample that was titrated and calculate the molarity in the boiler scale removal solution.

Explanation / Answer

Calculation of titration result is always based on the stoichiometry of the titration reaction.

When titrating we are usually given information about the volume and the concentration of the titrant solution and about volume of the titrated substance solution. Balanced reaction equation shows ratio of number of moles of reacting substances, thus to be able to deal with titration results we have to be able to convert between volumes, concentrations and numbers of moles. All these conversions are based on the definition of a molar concentration:
titration-calculation,

which can be rearranged to calculate number of moles when concentration and volume are given:
titration-calculation,

(where n is number of moles, C concentration and V is volume).

Write the chemical equation for the reaction:

1)HCl + NaOH ---> NaCl + H2O

2) The key molar ratio . . . :

. . . is that of HCl to NaOH, a 1 to 1 ratio.

3) Determine moles of HCl:

1 is to 1 as 0.000100 mol is to x

x = 0.000100 mol of HCl consumed

4) Determine moles of NaOH:

moles = MV = (0.125 mol/L) (0.0100 L) = 0.00125 mol

5) Determine molarity of NaOH solution:

0.00125 mol / 0.04400 L = 0.02841 M

Solution #2 (specific to a 1:1 ratio):

M1V1 = M2V2

(0.125 mol/L) (10.00 mL) = (x) (44.00 mL)

x = 0.02841 M

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