Question #3b25a

Question

1520 views around the world

You can reuse this answer
Creative Commons License

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")#