100 mL 0.1 M HCl was mixed with 150 mL 0.1 M HNO3. What volume of 0.1 M NaOH will be needed for complete neutralisation of the mxiture?
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158 mL
158 mL of 0.1 M of HCl is required to react completely with 0.1 M mixture of NaOH, containing equimolar amounts of both.
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Given:
Molarity of all the compounds, M1 = M2 = M3 = 0.1 M
V1 = 100 ml
V2 = 150 ml
To Find:
The volume of NaOH needed for complete neutralization.
Calculation:
- Since HCl and HNO3 both are monobasic acids, hence we can simply add their volumes.
- Now using molarity equation, we get:
M1 × (V1 + V2) = M3 × V3
⇒ 0.1 (100 + 150) = 0.1 × V3
⇒ V3 = 250 ml
- So, we need 250 ml of NaOH for the complete neutralization of the mixture.
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