Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
[tex]\text{ Cu\lparen NO}_3)_2\text{ }\rightarrow\text{ Cu}^{2+}\text{ + 2NO}_3^-[/tex]
The NO3- ion is the conjugate base of a strong acid, it will not affecy the pH, but Cu2+ ion is not conjugate acid of a strong base, it will affect the pH.
[tex]\text{ Cu}^{2+}\text{ + 2H}_2\text{O }\rightleftarrows\text{ Cu\lparen OH\rparen}_2\text{ + 2H}^{+\text{ }}\text{ Ka=3.00x10}^{-8}^{\frac{}{}}[/tex][tex]\text{ Ka= }\frac{\lbrack\text{ H}^+\rbrack^2}{\lbrack\text{ Cu}^{2+}\rbrack}[/tex]we can use the approximation that [Cu2+] = initial concentration of Cu(NO3)2 because the is Ka is very small
Solve for [H+]:
[tex]\begin{gathered} \lbrack\text{ H}^+\rbrack\text{ = }\sqrt{\text{ Ka }\times\text{ }\lbrack\text{ Cu}^{2+}\rbrack} \\ \lbrack\text{ H}^+\rbrack\text{ = }\sqrt{3.00\times10^{-8}\text{ }\times\text{ 1.30}}=\text{ 1.97}\times10^{-4} \end{gathered}[/tex][tex]\begin{gathered} \text{ pH = }-\text{ log \lparen}\lbrack\text{H}^+\rbrack) \\ \text{ pH = - log \lparen1.97}\times10^{-4})\text{ = 3.71 } \end{gathered}[/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
