Answer:
pH = 10.9
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:
[tex]n_{CO_3^{2-}}=0.34mol/L*0.250L+0.0585mol=0.1435mol\\\\n_{HCO_3^{-}}=0.34mol/L*0.250L-0.0585mol=0.0265mol[/tex]
The resulting concentrations are:
[tex][CO_3^{2-}]=\frac{0.1435mol}{0.25L}=0.574M \\[/tex]
[tex][HCO_3^{-}]=\frac{0.0265mol}{0.25L}=0.106M[/tex]
Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:
[tex]pH=10.2+log(\frac{0.574M}{0.106M} )\\\\pH=10.9[/tex]
Which makes sense since basic OH⁻ ions were added.
Regards!
