Titration Problem - Calculate the concentration of 20.0 mL of an H2SO4 solution that requires 27.7 mL of a 0.100 M NaOH solution?

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Titration Problem - Calculate the concentration of 20.0 mL of an H2SO4 solution that requires 27.7 mL of a 0.100 M NaOH solution?

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Answer 1 · 2018-04-14T17:49:14

$0.06925 M$ $0.06925M$

Explanation:

$2 N a O H + H_{2} S O_{4}$ $2NaOH+H_2SO_4$ ---> $N a_{2} S O_{4} + 2 H_{2} O$ $Na_2SO_4+2H_2O$

First calculate the number of moles (or amount) of the known solution, which in this case is the NaOH solution.
The volume of NaOH is $27.7 m L$ $27.7 mL$ , or $0.0277 L$ $0.0277L$ .
The concentration of NaOH is $0.100 M$ $0.100M$ , or in other words, $0.100$ $0.100$ mol/L

Amount = concentration x volume

$0.0277 L \times 0.100 M = 0.00277$ $0.0277Lxx0.100M=0.00277$ $m o l$ $mol$

As you can see from the reaction equation, the amount of $H 2_{S} O_{4}$ $H2_SO_4$ is half the amount of $N a O H$ $NaOH$ , as there is $2 N a O H$ $2NaOH$ but only $1 H_{2} S O_{4}$ $1H_2SO_4$

Amount of $H_{2} S O_{4} = \frac{0.00277}{2} = 0.001385$ $H_2SO_4=0.00277/2=0.001385$ $m o l$ $mol$

Concentration = amount/volume

$0.001385$ $0.001385$ $m o l$ $mol$ / $0.02 L = 0.06925 M$ $0.02L=0.06925M$

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Titration Problem - Calculate the concentration of 20.0 mL of an H2SO4 solution that requires 27.7 mL of a 0.100 M NaOH solution?

1 Answer

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