| 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. \[\text{100mL}\frac{\text{M}}{\text{10}}\text{HCl+100mL}\frac{\text{M}}{\text{10}}\text{NaOH}\] |
A) D
B) A
C) B
D) C
Correct Answer: D
Solution :
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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