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How many mL pf 1.07 M HClO4 should be added to 1.00 g of imidazole (C3H4N2;...

How many mL pf 1.07 M HClO4 should be added to 1.00 g of imidazole (C3H4N2; MW= 68.077 g/mol), a weak organic base (pKb= 6.95), to give a pH of 7.993?

Expert Solution

Number of moles of imidazole(C3H4N2) in 1g can be calculated as follows:

Moles of C3H4N2 = 1.00 g C3H4N2 x (1 mole C3H4N2 / 68.077 g C3H4N2) = 0.0147 moles C3H4N2.

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Addition of strong acid HClO4 to weak bases imidazole (C3H4N2) creates a buffer. The moles of protonated imidazole(C3H4N2H+) formed in the buffer by addition of HClO4 can be calculated by using Henderson-Hasselbalch’s equation as follows:

pH = pKa + log (moles C3H4N2 / moles C3H4N2H+)

pH = (14 – pKb) + log (moles C3H4N2 / moles C3H4N2H+)

7.993 = (14 – 6.95) + log (moles C3H4N2 / moles C3H4N2H+)

7.993 = 7.05 + log (moles C3H4N2 / moles C3H4N2H+)

7.993 – 7.05 = log (moles C3H4N2 / moles C3H4N2H+)

0.943 = log (moles C3H4N2 / moles C3H4N2H+)

10^0.943 = (moles C3H4N2 / moles C3H4N2H+)

8.770 = (moles C3H4N2 / moles C3H4N2H+)

moles C3H4N2 = 8.770 (moles C3H4N2H+)-----------------(1)

Since, initial amount of (C3H4N2) = 0.0147 = (moles C3H4N2) + (moles C3H4N2H+)

Therefore, (moles C3H4N2H+) = 0.0147 - (moles C3H4N2)

Substitute eq 1 in above equation

(moles C3H4N2H+) = 0.0147 - (8.770 (moles C3H4N2H+))

(moles C3H4N2H+) + (8.770 (moles C3H4N2H+))= 0.0147

9.770 (moles C3H4N2H+) = 0.0147

(moles C3H4N2H+) = 0.0147 / 9.770 = 0.00150

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Since, moles of C3H4N2H+ are formed by addition of HClO4.

Moles of HClO4 = moles of C3H4N2H+ = 0.00150

Molarity of HClO4 is 1.07M, hence, volume of HClO4 can be calculated as follows:

Volume of HClO4(L) = moles of HClO4 / molarity of HClO4

Volume of HClO4(L) = 0.00150 / 1.07

Volume of HClO4(L) = 0.00140 L = 1.4mL

Hence, 1.4 mL of 1.07 M HClO4 should be added to 1.00 g of imidazole to get a pH of 7.993

queen_honey_blossom answered 8 months ago

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