Question

What volume of 0.10 molarity h2so4 must be added to 50 ml of 0.10 molarity and naoh solution to make a solution in which the molarity of the h2so4 is 0.050 molar arity?

Answer 1

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Answer 2

Answer: The total volume of $H_2SO_4$ that must be added is 100 mL.

Explanation:

To calculate the volume of acid, 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:

$n_1=2\\M_1=0.10M\\V_1=?mL\\n_2=1\\M_2=0.10M\\V_2=50mL$

Putting values in above equation, we get:

$2\times 0.10\times V_1=1\times 0.10\times 50\\\\V_1=25mL$

Let the volume of sulfuric acid that must be added more is 'x' mL

So, the total volume of the solution becomes = (25 + 50)mL = 75 mL

We are now diluting the solution with some amount of sulfuric acid.

• To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=0.10M\\V_1=xmL\\M_2=0.05\\V_2=(75+x)mL$

Putting values in above equation, we get:

$0.10\times x=0.05\times (75+x)\\\\x=75mL$

Thus, the total volume of $H_2SO_4$ that is added in the solution = (25 + 75)mL = 100 mL

Hence, the total volume of $H_2SO_4$ that must be added is 100 mL.