Answer:
For 1: The molarity of acidic solution is 0.579 M
For 2: The volume of acidic solution is 0.0153 L
Explanation:
To calculate the concentration of acid, 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
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base
We are given:
[tex]n_1=1\\M_1=?M\\V_1=871mL\\n_2=3\\M_2=0.030M\\V_2=560mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 871=3\times 0.0300\times 560\\\\M_1=\frac{3\times 0.0300\times 560}{1\times 871}=0.579M[/tex]
Hence, the molarity of acidic solution is 0.579 M
We are given:
[tex]n_1=3\\M_1=0.488M\\V_1=?L\\n_2=2\\M_2=0.320M\\V_2=35.1mL=0.0351L[/tex]
Putting values in above equation, we get:
[tex]3\times 0.488\times V_1=2\times 0.320\times 0.0351\\\\V_1=\frac{2\times 0.320\times 0.0351}{3\times 0.488}=0.0153L[/tex]
Hence, the volume of acidic solution is 0.0153 L