A) \[55mL\frac{M}{10}HCl+45mL\frac{M}{10}NaOH\]
B) \[75mL\frac{M}{5}HCl+25mL\frac{M}{5}NaOH\]
C) \[\text{100mL}\frac{\text{M}}{\text{10}}\text{HCl+100mL}\frac{\text{M}}{\text{10}}\text{NaOH}\]
D) \[\text{60mL}\frac{M}{10}HCl+40mL\frac{M}{10}NaOH\]
Correct Answer: B
Solution :
\[75mL\frac{M}{5}HCl+25mL\frac{M}{5}NaOH\] \[25mL\frac{M}{5}NaOH\] will neutralise \[25mL\frac{M}{5}HCl\] \[75-25=50mL\frac{M}{5}HCl\] will remain. Total volume will be \[75+25=100mL.\] \[[{{H}^{+}}]=[HCl]=\frac{M}{5}\times \frac{50}{100}=\frac{M}{10}\] \[pH=-{{\log }_{10}}[{{H}^{+}}]=-{{\log }_{10}}\frac{M}{10}=1\]
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