A chemistry student weighs out 0.09666 g of phosphoric acid (H3PO4), a triprotic acid, into a 250.volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaoH solution. Calculate the volume of NaoH solution the student will need to add to reach the final equivalence point. Round your answer to 3 significant digits.

Answer :

Answer: 14.62 ml

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n= moles of solute  

[tex]V_s[/tex] = volume of solution in ml = 250 ml

[tex]{\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{0.09666g}{98g/mol}=9.9\times 10^{-4}[/tex]

Now put all the given values in the formula of molarity, we get

[tex]Molarity=\frac{9.9\times 10^{-4}\times 1000}{250ml}[/tex]

[tex]Molarity=3.9\times 10^{-3}M[/tex]

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

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_3PO_4[/tex]

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

We are given:

[tex]n_1=3\\M_1=3.9\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.2000M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]3\times 3.9\times 10^{-3}\times 250=1\times 0.2000\times V_2\\\\V_2=14.62ml[/tex]

Thus the volume of NaOH solution the student will need to add to reach the final equivalence point is 14.62 ml

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