Question #3b25a

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Answer 1 · 2016-12-02T01:09:58

$sf([H^+]=5.34xx10^(-12)color(white)(x)"mol/l")$

$sf(NH_3+H_2OrightleftharpoonsNH_4^(+)+OH^-)$

For a weak base like this we can use the expression:

$sf(pOH=1/2(pK_b-log[base])$

Putting in the numbers:

$sf(pOH=1/2(5.00-log(0.35))$

$:.$ $sf(pOH=1/2xx5.4559=2.727)$

$sf(pOH+pH=14)$

$:.$ $sf(pH=14-pOH=14-2.727=11.27)$

Since $sf([H^+]=10^(-pH))$

This gives:

$sf([H^+]=5.34xx10^(-12)color(white)(x)"mol/l")$