QUESTIONS A student was required to prepare 260.0 mL of a dichloroacetio acid/so

Question

QUESTIONS A student was required to prepare 260.0 mL of a dichloroacetio acid/sodum M. The student was supplied with 0.764 M dichlonoacetic acid and 1.0M NaOH to perform this task. What volume (in L) of the acid would the student need to prepare this buffler solution bufer in which the concentration of the weak acid component was 0,.07 M and the corcentration of the hat all of the oonjugate base comes directly from the maction of NaOH with the weak acid (in other words, there is negigible dissociation of the weak acid

Explanation / Answer

In this buffer.

Moles of acid = 0.25*0.07 = 0.0175

Moles of conjugate base = 0.25*0.034 = 0.0085

Let's assume that we take 'x' L of the stock solution of acid and 'y' L of NaOH stock solution initially

Moles of acid initially = 0.764*x

Moles of NaOH initially = 1*y = y

So,

Moles of conjugate base = Moles of NaOH taken = y = 0.0085

0.764*x - y = 0.0175

So,

0.0764*x - 0.0085 = 0.0175

Solving we get:

x = 0.34

So,

Volume of acid stock needed = 0.34 L

Volume of NaOH needed = 0.0085 L

Hope this helps !

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