Respuesta :
the balanced equation for the above reaction is as follows;
NaOH + HCl -- > NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of HCl moles reacted - 0.0100 mol/L x 0.02060 L = 0.000206 mol
according to 1:1 molar ratio
number of NaOH moles reacted - 0.000206 mol
number of NaOH moles in 30.00 mL - 0.000206 mol
therefore number of NaOH moles in 1000 mL - 0.000206 mol / 0.03000 L = 0.00687 mol
molarity of NaOH is 0.00687 M
NaOH + HCl -- > NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of HCl moles reacted - 0.0100 mol/L x 0.02060 L = 0.000206 mol
according to 1:1 molar ratio
number of NaOH moles reacted - 0.000206 mol
number of NaOH moles in 30.00 mL - 0.000206 mol
therefore number of NaOH moles in 1000 mL - 0.000206 mol / 0.03000 L = 0.00687 mol
molarity of NaOH is 0.00687 M
Answer : The molarity of the NaOH solution is, 0.00687 M.
Explanation :
The balanced chemical reaction will be:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
Using neutralization method :
[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=0.0100M\\V_1=20.60mL\\n_2=1\\M_2=?\\V_2=30.00mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.0100M\times 20.60mL=1\times M_2\times 30.00mL\\\\M_2=0.00687M[/tex]
Hence, the molarity of the NaOH solution is, 0.00687 M.
