Answer :
Answer:
0.013 M NaOH
Step-by-step explanation:
HCL + NaOH = NaCL + water
The mole ratio of acid and base is 1:1
Molarity is equal to the moles of solute divided by the liters of solution
Solute mole = 12.5 × 10^-3 × 0.3
= 0.003075
To neutralize 0.00375mol of HCl, 0.00375 mol of NaOH will be required. Since molarity is moles per liter, divide this amount of moles by the volume in liters:
Molarity = 0.00375/285 × 10^-3
Molarity = 0.0131 M
The molarity concentration of the NaOH solution is 0.013M NaOH
The equation of reaction is given as:
[tex]HCl + NaOH >>>>>> NaCl + H_2O[/tex]
Concentration of Hcl, [tex]C_{Hcl}=0.30M[/tex]
Volume of Hcl, [tex]V_{HCl}=12.5mL[/tex]
Concentration of NaOH = [tex]C_{NaOH}[/tex]
Volume of NaOH, [tex]V_NaOH = 285mL[/tex]
From the equation of reaction:
Number of moles of Hcl, [tex]n_{HCl}=1[/tex]
Number of moles of NaOH, [tex]n_{NaOH}=1[/tex]
The formula for neutralization is given as:
[tex]\frac{C_{HCl}V_{HCl}}{C_{NaOH}V_{NaOH}} = \frac{n_{HCl}}{n_{NaOH}}[/tex]
Substitute the values of the parameters
[tex]\frac{0.30(12.5)}{C_{NaOH}(285)} =\frac{1}{1} \\\\C_{NaOH}(285) = 0.30(12.5)\\\\C_{NaOH}(285) = 3.75\\\\C_{NaOH}=\frac{3.75}{285} \\\\C_{NaOH} = 0.013M[/tex]
The molarity concentration of the NaOH solution is 0.013M NaOH
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