What volume (to the nearest 0.1 mL) of 6.00-M NaOH must be added to 0.300 L of 0

Question

What volume (to the nearest 0.1 mL) of 6.00-M NaOH must be added to 0.300 L of 0.350-M HNO2 to prepare a pH = 3.00 buffer?

Explanation / Answer

This would be a two step solution: a stoichiometric part and an equilibrium: First the Stoich (in moles): _____HNO2 + OH- -> H2O + NO2- Initial__.045___x_____________0 Final__.045-x__0_____________x Where x is the number of moles of NaOH that need to be added. H2O is not considered here because it is not important at all. Now for the Equilibrium part (the dissociation of HNO2): but first we have to convert back to molarity instead of moles (because we don't know the total volume, it would be safe to assume 0.3 L as the total because the molarity of NaOH is so high). And we know we want a pH of 3.60, so we can determine the final [H+] concentration, which is 2.512*10^-4 ________HNO2 H+ + NO2- Initial___(.045-x)/.3____0_____x/.3 Change____-y_______+y_____+y Equil.___0.15-x/.3 -y___y____x/.3 + y So therefore, y=2.512*10^-4 Ka of HNO2 = 4.0*10^-4 Then writing out the equilibrium expression: 4.0*10^-4 = [H+][NO2-]/[HNO2] 4.0*10^-4 = (2.512*10^-4)(x/.3 + 2.512*10^-4)/(.15 - x/.3 - 2.512*10^-4) Then to solve for x, I hope I don't have to show the algebra, that'd take forever! Well, I solved this graphically, and I found x=.02756 x is the number of moles of NaOH 0.02756 mol * 1L/7.5mol = .00368L = 3.68mL which is roughly equal to 3.7mL ANSWER (after that long work): 3.7mL

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