Question #873de

1 Answer
Jul 31, 2016

Answer:

#sf(9.65)#

Explanation:

The #sf(HCO_3^-)# ion is basic:

#sf(HCO_(3(aq))^(-)+H_2O_((l))rightleftharpoonsH_2CO_(3(aq))+OH_((aq))^-)#

#sf(K_(b)=([H_2CO_3][OH^(-)])/([HCO_3^(-)])=1.99xx10^(-8)color(white)(x)"mol/l")#

We know that #sf([H_2CO_3]=[OH^(-)])#

Because the dissociation is small I will assume that the equilibrium concentration of #sf([HCO_3^-]}# is equal to #sf(0.1color(white)(x)"mol/l")#

So now we can write:

#sf(K_b=[OH^(-)]^(2)/(0.1)=1.99xx10^(-8))#

#:.##sf([OH^-]^(2)=1.99xx10^(-8)xx0.1=1.99xx10^(-9))#

From which:

#sf([OH^-]=4.46x10^(-5)color(white)(x)"mol/l")#

#:.##sf(pOH=-log(4.46xx10^(-5))=4.35)#

#sf(pH+pOH=14)#

#:.##sf(pH=14-4.35=9.65)#