Question #ab631

1 Answer
Mar 3, 2017

Answer:

You can do it like this:

Explanation:

If you have a weak acid then you can use this expression:

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

Where a is the concentration of the acid. We assume that it is a close approximation to the equilibrium concentration.

#sf(M_r=204.22)#

#:.##sf(n=m/M_r=0.180/204.22=8.814xx10^(-4)color(white)(x)"mol")#

#sf(c=n/v=(0.180xx10^(-4))/(50.0/1000)=0.01763color(white)(x)"mol/l")#

#:.##sf(pH=1/2[5.4-log(0.01763)])#

#sf(pH=1/2xx7.154=3.58)#