Respuesta :
Answer : The concentration of the [tex]Ba(OH)_2[/tex] solution is, 2.16 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]HCOOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex].
We are given:
[tex]n_1=1\\M_1=1.120M\\V_1=31.4mL\\n_2=2\\M_2=?\\V_2=16.3mL[/tex]
Putting values in above equation, we get:
[tex]1\times 1.120M\times 31.4mL=2\times M_2\times 16.3mL\\\\M_2=2.16M[/tex]
Thus, the concentration of the [tex]Ba(OH)_2[/tex] solution is, 2.16 M
the concentration of the Ba(OH)2 solution is 2.16 M.
Calculation of the concentration:
the following equation should be used
n1M1v1 = n2M2V2
n1M1v1 are the n-factor, molarity and volume of acid
n2M2V2 are the n-factor, molarity and volume of base
Since
n1 = 1
M1 = 1.120 M
V1 = 31.4 mL
n2 = 2
V2 = 16.3 mL
So, the concentration should be
1*1.120*31.4 = 2* m2*16.3
m2 = 2.16 M
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