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In carrying out the first standardization in this experiment, a student used 0.5793 g of potassium hydrogen phthalate (khp) and 24.55 ml of a naoh solution are needed to reach the equivalence point. what is the concentration (in mol/l) of the student's naoh solution? include only the numerical answer (no units).



Answer :

In carrying out the first standardization in this experiment, a student used 0.5793 g of potassium hydrogen phthalate (khp) and 24.55 ml of a NaOH solution are needed to reach the equivalence point then the concentration of student NaOH solution is 0.02455 L

Standardization is the process in which determining the exact concentration of molarity of the solution and here given data is

Mass of potassium hydrogen phthalate (C₈H₅KO₄) = 0.5793 g

Molar mass of potassium hydrogen phthalate (C₈H₅KO₄) = 204.22g/mol

The number of mole of potassium hydrogen phthalate present then,

Number of moles = mass/molar mass

Number of moles = 0.5793 g/204.22g/mol = 0.002836 mole

Volume of NaOH present = 24.55 ml

1 ml =  0.001 L then, 24.55 ml = 24.55 ml×0.001 L = 0.02455 L

The concentration (in mol/l) of the student's naoh solution is 0.02455 L

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