Respuesta :
Answer:
1.692 M is the molarity of the unknown solution of HCl.
Explanation:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
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 which is [tex]HCL[/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=1\\M_1=?\\V_1=19.50 mL\\n_2=1\\M_2=2.750 M\\V_2=12.00 mL[/tex]
Putting values in above equation, we get:
[tex]M_1=\frac{n_2M_2V_2}{n_1V_1}[/tex]
[tex]=\frac{1\times 2.750 M\times 12.00 mL}{1\times 19.50 mL}[/tex]
[tex]M_1=1.692 M[/tex]
1.692 M is the molarity of the unknown solution of HCl.
Answer:
4.46875 M.
Explanation:
Equation of the reaction
NaOH + HCl --> NaCl + H2O
Molar concentration = number of moles/volume
Number of moles = 2.75 × 0.0195
= 0.0536 mol
By stoichiometry, 1 mole of HCl reacted with 1 mole of NaOH.
Therefore,
Number of moles of HCl = 0.0536 mol.
Molar concentration in 12 ml = 0.0536/0.012
= 4.46876 M.
