| Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations : [NEET - 2018] |
| [a] \[\text{60mL}\frac{\text{M}}{\text{10}}\text{HCl+40mL}\frac{\text{M}}{\text{10}}\text{NaOH}\] |
| [b] \[\text{55mL}\frac{\text{M}}{\text{10}}\text{HCl+45mL}\frac{\text{M}}{\text{10}}\text{NaOH}\] |
| [c] \[\text{75mL}\frac{\text{M}}{\text{5}}\text{HCl+25mL}\frac{\text{M}}{\text{5}}\text{NaOH}\] |
| [d] \[\text{100mL}\frac{\text{M}}{\text{10}}\text{HCl+100mL}\frac{\text{M}}{\text{10}}\text{NaOH}\] |
| pH of which one of them will be equal to 1? |
A) d
B) a
C) b
D) c
Correct Answer: D
Solution :
| [d] |
| Meq of \[\text{HCl=75}\times \frac{1}{5}\times 1=15\] |
| Meq of \[\text{NaOH=25 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{5}}\text{ }\!\!\times\!\!\text{ 1=5}\] |
| Meq of HCl in resulting solution =10 |
| Molarity of \[\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ }\] in resulting mixture |
| \[\text{=}\frac{10}{100}=\frac{1}{10}\] |
| \[\text{pH=-log }\!\![\!\!\text{ }{{\text{H}}^{+}}]=-\log \left[ \frac{1}{10} \right]=1.0\] |
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