The methyl red indicator you used in this titration is actually a weak acid that

Question

The methyl red indicator you used in this titration is actually a weak acid that we can abbreviate as "HM". The methyl red participates in the following weak acid equilibrium: HM (aq) Doubleheadarrow H^+ (aq) + M^-(aq) K_a = [H^+][M^-]/[HM] The indicator is present in such a small quantity that it does not affect the pH of the solution. Instead, the pH of the solution determines which form of methyl red (HM or M^-) is predominantly present. The color of HM and M^- are different, which accounts for the color change of the solution. Answer the following questions based on your observations during this titration: What is the color of HM? What is the color of M^-? C. Calculate (or, more accurately, roughly estimate) the pK_a and the K_a of the methyl red indicator.

Explanation / Answer

HM (aq) -----> H+ (aq) + M- (aq)

Ka = [H+] [M-] / [HM] -------- eq 1

pKa = pH + log10 [M-] / [HM]

  pKa - pH = log10 [M-] / [HM]

When pH becomes equal to pKa then HM will also become equal to M-.

Similarly, we can also write, pH = pKa + log10 [M-] / [HM]

So, pH = pH + log10 [M-] / [M-]

Log10 (1) = 0, (when pH = pKa & M- = HM)

From eq 1,

Ka = [H+] [M-] / [HM]

= [H+] [M-] / [M-]

Thus, Ka = [H+]

  pKa = pH + log10 [M-] / [HM]

At this point, the solution will become orange in colour.

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