A) \[3.2\]
B) \[3.85\]
C) \[10.15\]
D) None of these
Correct Answer: C
Solution :
[c] 88 ppm means: \[{{10}^{6}}\text{ }mL\]of air contains \[=88\text{ }g\]of \[C{{O}_{2}}\] \[=\frac{88}{44}=2\,molC{{O}_{2}}\] Moles of \[C{{O}_{2}}\] is \[{{10}^{5}}mL=\frac{2}{{{10}^{6}}}\times {{10}^{5}}=0.2mol\] \[[C{{O}_{2}}]=\frac{0.2mol}{1d{{m}^{3}}}=0.2M\] Use direct reaction \[pH{{w}_{A}}=\frac{1}{2}(p{{K}_{a}}-\log C)\] \[=\frac{1}{2}(7-\log \,\,0.2)\] \[=\frac{1}{2}(7-0.3+1)\] \[=\frac{7.7}{2}\] \[=3.85\] \[=pOH=14-3.85=10.15\]You need to login to perform this action.
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