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If we plan to prepare a buffer with the $\mathrm{pH}$ of $7.35$ using $\ce{HClO}$ ($\mathrm pK_\mathrm a = 7.54$), what mass of the solid sodium salt of the conjugate base is needed to make this buffer? Suppose you want to use $\pu{125.0mL}$ of $\pu{0.500M}$ of the acid.

I know this relates to Henderson's equation, so I do:

$$7.35=7.54+\log{\frac{[\ce{ClO-}]}{[\ce{HClO}]}},$$

which becomes:

$$0.646=\frac{[\ce{ClO-}]}{[\ce{HClO}]}.$$

But I do not know how to go from there, and I don't know how to use the last piece of information in the problem: ("Suppose you want to use $\pu{125.0mL}$ of $\pu{0.500M}$ of the acid").

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  • $\begingroup$ Hello and welcome to the Chemistry.SE! You should take the tour to get to know our community and learn how to ask questions. As it stands now, you title is way too long and unclear. Try to state one clearly formulated question. $\endgroup$ – Fl.pf. Apr 7 '17 at 7:30
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The information given in the problem, "Suppose you want to use 125.0mL of 0.500M of the acid." tells us that the molarity or concentration of the acid is 0.5M. So, [ACID] = 0.5. Now, 0.646 = [BASE]/(0.5) So, [BASE] = 0.646×0.5 = 0.323 So, concentration of conjugate base = 0.323M n/V = 0.323 Since, volume is 125.0mL = 0.125L n/(0.125) = 0.323 So, n = 0.04 So, no. of moles of conjugate base = 0.04 So, mass of sodium salt of conjugate base i.e NaClO = 0.04×74.5 ~= 3g (Since, molar​ mass of NaClO is 74.5) Thus, your answer is 3g.

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