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Determine the ph of a 0.22 m naf solution at 25° c. the salt completely dissociates into na (aq) and f- (aq), and the na (aq) ion has not acid or base properties. the ka of hf is 3.5 × 10-5.

Ramon Kelly

in Chemistry

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Heather Maxwell on February 5, 2019

Doing the ICE approache naf ---> na + f-Initial 0.22 Change -0.22 0.22 0.22 Balance 0 0.22 0.22 Hydrolysis will continue, the dissociation f- + H2O ----> hf + OH-Initial 0.22 Change -x x xEquilibrium 0.22-x x xWe are given with the ka of hfhf = 3.5x10^-5 = x(x)/(0.22-x)of the Solution to that xx = 2.76x10^-3 m, which is the concentration of OH - ionsSolving for pHpH = 14 - pOHpH = 14 - (-log [2.76x10^-3])pH = 11.44 pH is 11.44 which means the solution is basic.


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