One liter of 0.400 M NH3 solution also contains 12.78 g of NH4Cl. What is the pH of this buffer?

a. What is the pH of this buffer if 0.142 mol of HCl is bubbled through the solution?

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  • 4 months ago

    Moles NH4Cl = 12.78 g / 53.491 g/mol = 0.2389 mol NH4+

    Ka for NH4+ = 5.6 x 10^-10. pKa = 9.25

    So, the initial solution contains 0.400 moles of NH3 and 0.2389 moles of NH4+

    pH = pKa + log [NH3]/[NH4+]

    pH = 9.25 + log (0.400 / 0.2389) = 9.47

    a. Adding 0.142 moles HCl quantitatively protonates 0.142 mol NH3 forming an additional 0.142 mol NH4+. So, after this addition, moles NH3 = 0.400 - 0.142 = 0.258, and moles NH4+ - 0.2389 + 0.142 = 0.381 

    pH = 9.25 + log (0.258 / 0.381) = 9.08

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