What volume of 0.01 M $H_{2} S O_{4}$ must be added to 50ml of 0.01M NaOH solution to make a solution in which the molarity of the $H_{2} S O_{4}$ is 0.050M?

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Solution

Normality of $H_{2} S O_{4} = M_{1} = 0.05 M$
Normality of $N a O H = M_{2} = 0.1 M$
Vol. of NaOH $= V_{2} = 50 m l$
Vol. of $H_{2} S O_{4} = V_{1} = ?$
Now $M_{1} V_{1} = M_{2} V_{2}$
$V_{1} = \frac{M_{2} V_{2}}{M_{1}}$
$= \frac{0.1 \times 50}{0.05}$
Vol. of $H_{2} S O_{4} = V_{1} = 100 m l$

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What volume of 0.01 M H2SO4 must be added to 50ml of 0.01M NaOH solution to make a solution in which the molarity of the H2SO4 is 0.050M?

Normality of H2SO4=M1=0.05MNormality of NaOH=M2=0.1MVol. of NaOH =V2=50mlVol. of H2SO4=V1=?Now M1V1=M2V2V1=M2V2M1=0.1×500.05Vol. of H2SO4=V1=100ml

What volume of 0.01 M $H_{2} S O_{4}$ must be added to 50ml of 0.01M NaOH solution to make a solution in which the molarity of the $H_{2} S O_{4}$ is 0.050M?

Normality of $H_{2} S O_{4} = M_{1} = 0.05 M$
Normality of $N a O H = M_{2} = 0.1 M$
Vol. of NaOH $= V_{2} = 50 m l$
Vol. of $H_{2} S O_{4} = V_{1} = ?$
Now $M_{1} V_{1} = M_{2} V_{2}$
$V_{1} = \frac{M_{2} V_{2}}{M_{1}}$
$= \frac{0.1 \times 50}{0.05}$
Vol. of $H_{2} S O_{4} = V_{1} = 100 m l$