Answer: The molarity of base is 0.103 M
Explanation:
To calculate the concentration 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]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]NH_4OH[/tex]
We are given:
[tex]n_1=1\\M_1=0.175M\\V_1=29.50mL\\n_2=1\\M_2=?M\\V_2=50.0mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.175\times 29.50=1\times M_2\times 50.0\\\\M_2=\frac{1\times 0.175\times 29.50}{1\times 50.0}=0.103M[/tex]
Hence, the molarity of base is 0.103 M
