%3Cp%3E5%20mg%26nbsp%3Bof%26nbsp%3BHOCL%26nbsp%3Bis%26nbsp%3Badded%26nbsp%3Bto%2

Question

%3Cp%3E5%20mg%26nbsp%3Bof%26nbsp%3BHOCL%26nbsp%3Bis%26nbsp%3Badded%26nbsp%3Bto%26nbsp%3Bpure%26nbsp%3Bwater%2C%26nbsp%3Bsuch%26nbsp%3Bthat%26nbsp%3Bthe%26nbsp%3Bfinal%26nbsp%3Bvolume%26nbsp%3Bis%26nbsp%3B1L.%26nbsp%3BCalculate%26nbsp%3Bthe%26nbsp%3BpH%26nbsp%3Bat%26nbsp%3Bwhich%26nbsp%3B75%25%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3BHOCL%26nbsp%3Badded%26nbsp%3Bis%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Bprotonated%26nbsp%3Bform%26nbsp%3B(HOCL)%26nbsp%3Band%26nbsp%3B25%25%26nbsp%3Bis%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Bdeprotonated%26nbsp%3Bform%26nbsp%3B(OCL-).%3C%2Fp%3E

Explanation / Answer

5 mg of HOCL is added to pure water, such that the final volume is 1L. Calculate the pH at which 75% of the HOCL added is in the protonated form (HOCL) and 25% is in the deprotonated form (OCL-).

[HOCL] = ?

moles of HOCl = 5*10^-3 /52.46 = 9.53*10^-5

[HOCl] = moles/ volume = 9.53*10^-5

HOCl <=> H+ and OCl-

[H+] = 0.25*[HOCl] = 2.3827*10^-5

pH = - log [H+] = 4.623

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