A) \[7.583\times {{10}^{-5}}\]
B) \[6.583\times {{10}^{-5}}\]
C) \[7.03\times {{10}^{-5}}\]
D) \[7.583\times {{10}^{-3}}\]
Correct Answer: D
Solution :
\[{{E}_{cell}}=-0.3665\,V\] \[{{E}_{SHE}}=0.2412\,V\] \[{{E}_{cell}}=-0.2412+x=-0.3665\] \[x=-0.3665+0.2412\] \[=-0.1253\] For hydrogen half-cell: \[-0.1253=\frac{0.0591}{2}\log {{[{{H}^{+}}]}^{2}}\] \[\log {{[{{H}^{+}}]}^{2}}=-\frac{0.1253\times 2}{0.0591}=-4.2402\] \[[{{H}^{+}}]=7.583\times {{10}^{-3}}M\] The hydrogen half-cell is a buffer of\[{{H}_{3}}P{{O}_{4}}\] and\[{{H}_{2}}PO_{4}^{-}\]. \[\therefore \] \[[{{H}^{+}}]={{K}_{{{a}_{1}}}}\frac{[acid]}{[salt]}\] \[\frac{[acid]}{[salt]}=1\] \[[{{H}^{+}}]={{K}_{{{a}_{1}}}}\] \[{{K}_{{{a}_{1}}}}=7.583\times {{10}^{-3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
