Rhubarb leaves contain oxalic acid, a toxic substance that...

Rhubarb leaves contain oxalic acid, a toxic substance that can cause stomach irritation and kidney problems. Oxalic acid solutions reach equilibrium with both of these reactions occurring. Reaction 1: HOOCCOOH(aq) + H$_2$O(l) = H$_3$O$^+$(aq) + HOOCCOO$^-$(aq) hydrogen oxalate ion oxalic acid Reaction 2: HOOCCOO$^-$(aq) + H$_2$O(l) = H$_3$O$^+$(aq) + OOCCOO$^{2-}$(aq) oxalate ion 17. The substances that act as Bronsted-Lowry bases in reaction 2 are ? HOOCCOO$^-$(aq) and H$_2$O(l) ? HOOCCOO$^-$(aq) and OOCCOO$^{2-}$(aq) ? H$_2$O(l) and H$_3$O$^+$(aq) ? H$_2$O(l) and OOCCOO$^{2-}$(aq) 18. The substance in the above equilibria that is amphiprotic is ? HOOCCOO$^-$(aq) ? H$_3$O$^+$(aq) ? OOCCOO$^{2-}$(aq) ? HOOCCOOOH (aq)

Transcript

00:01 Hello students, given data in this question pka1 equal 1 .23, pka2 equal 4 .19 are given for diprotic oxalic acid.

00:10 Molarity of oxalic acid given is 0 .0349 molar.

00:14 The objective of this question is to find the ph of the given solution.

00:20 So let's calculate the hydronium ion concentration from pka1 and pka2 first to find the ph.

00:29 Ionization of oxalic acid gives h2c2o4 plus h2o ionizes to h3o plus plus hc2o4 minus.

00:38 Initial concentration given is 0 .0349 molar.

00:43 So at equilibrium concentration will be 0 .0349 minus x.

00:49 Now equilibrium constant is given by ka1 equals h3o plus concentration into hc2o4 minus concentration divided by h2c2o4 concentration.

01:02 Now substituting the values 1 .23 equals x square divided by, so here concentration of this is x and concentration of this is also x.

01:15 So this will become x square and concentration of this is 0 .0349 minus x.

01:23 So x square by 0 .0349 minus x.

01:29 So this implies 1 .23 into 0 .0349 minus 1 .23x equal to x square.

01:40 So this implies x square plus 1 .23x minus 0 .0429 equal to 0.

01:54 Now upon putting quadratic formula that is x equal to minus b plus or minus square root of b square minus 4ac by 2a.

02:04 So in this expression a is 1, b is 1 .23 and c is minus 0 .0429.

02:10 Now substituting we have x equals minus 1 .23 plus square root of 1 .23 square minus 4 into 1 into minus 0 .0429 divided by 2 into 1.

02:34 So we have to solve for x.

02:35 So on solving we get x equal to 0 .033.

02:43 Now so we got h3o plus concentration equals x that is 0 .0339 molar.

02:56 Now ionization of oxalic acid gives h2c2o4 minus plus h2o ionizing to h3o plus plus c2o4 minus.

03:07 Initial concentration is 0 .0339 molar and at equilibrium concentration becomes 0 .0339 minus y.

03:19 Now equilibrium constant ka2 equals h3o plus concentration into c2o4 2 minus concentration divided by hc2o4 minus concentration...

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