Question #89bee

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Answer 1 · 2017-07-09T21:38:25

#sf(K_a=1.08xx10^(-6))#

For a weak monoprotic acid this expression can be used:

#sf(pH=1/2(pK_a-log[acid])#

#:.##sf(2.9=1/2(pK_a-log1.46)#

#sf(2.9=1/2pK_a-1/2xx0.1643)#

#sf(1/2pK_a=2.98214)#

#sf(pK_a=5.9642)#

From which:

#sf(K_a=1.08xx10^(-6)color(white)(x)"mol/l")#