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Early people observed their surroundings and relied on...

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Damm [24]3 years ago

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In a neutralization reaction, 24.6 mL of 0.300 M H2SO4(aq) reacts completely with 20.0 mL of NaOH(aq). The products are Na2SO4(a

stiv31 [10]

Answer : The concentration of the NaOH solution is, 0.738 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL$

$M_2=0.738M$

Thus, the concentration of the NaOH solution is, 0.738 M

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