JEE MAIN - Chemistry (2003 - No. 1)
Which one of the following statements is not true?
pH + pOH = 14 for all aqueous solutions
The pH of 1 $$\times$$ 10-8 M HCl is 8
96,500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent of copper at the cathode
The conjugate base of $$H_2PO_4^-$$ is $$HPO_4^{2-}$$
Explanation
$$pH$$ of an acidic solution should be less than $$7.$$ The reason is that from $${H_2}O.\left[ {{H^ + }} \right] = {10^{ - 7}}M$$ which cannot be neglected in comparison to $${10^{ - 8}}M.$$ The $$pH$$ can be calculated as.
from acid, $$\,\,\,\left[ {{H^ + }} \right] = {10^{ - 8}}M.$$
from $${H_2}O,\,\,\left[ {{H^ + }} \right] = {10^{ - 7}}M$$
$$\therefore$$ $$\,\,\,$$ Total $$\left[ {{H^ + }} \right] = {10^{ - 8}} + {10^{ - 7}}$$
$$ = {10^{ - 8}}\left( {1 + 10} \right) = 11 \times {10^{ - 8}}$$
$$\therefore$$ $$\,\,\,pH = - \log \left[ {{H^ + }} \right]$$
$$ = - \log 11 \times {10^{ - 8}}$$
$$ = - \left[ {\log 11 + 8\log \,10} \right]$$
$$ = - \left[ {1.0414 - 8} \right] = 6.9586$$
from acid, $$\,\,\,\left[ {{H^ + }} \right] = {10^{ - 8}}M.$$
from $${H_2}O,\,\,\left[ {{H^ + }} \right] = {10^{ - 7}}M$$
$$\therefore$$ $$\,\,\,$$ Total $$\left[ {{H^ + }} \right] = {10^{ - 8}} + {10^{ - 7}}$$
$$ = {10^{ - 8}}\left( {1 + 10} \right) = 11 \times {10^{ - 8}}$$
$$\therefore$$ $$\,\,\,pH = - \log \left[ {{H^ + }} \right]$$
$$ = - \log 11 \times {10^{ - 8}}$$
$$ = - \left[ {\log 11 + 8\log \,10} \right]$$
$$ = - \left[ {1.0414 - 8} \right] = 6.9586$$
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