Question #873de

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Answer 1 · 2016-07-31T16:46:57

$sf(9.65)$

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:

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

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

$sf(pH+pOH=14)$

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