A) \[{{10}^{-pH}}+{{10}^{-pOH}}={{10}^{-14}}\]
B) \[pH\alpha \frac{1}{[{{H}^{+}}]}\]
C) \[{{K}_{w}}\alpha \,T\]
D) dissociation constant of water \[K=1.8\times {{10}^{-16}}\]
Correct Answer: A
Solution :
| [a] |
| [a] \[{{10}^{-pH}}\times {{10}^{-pOH}}={{10}^{-Kw}}\] |
| [b] as \[[{{H}^{+}}]\] increases, pH decreases, |
| [c] since ionization of water is endothermic in nature hence increase in temperature increases \[{{K}_{w}}\] |
| [d] \[{{K}_{a}}\] of |
| \[{{H}_{2}}O=\frac{[{{H}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}=\frac{{{10}^{-14}}}{55.55}=1.8\times {{10}^{-16}}\] |
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