Answer to Question #339545 in General Chemistry for Mary

Question

5g of NaOH and NaCL ( as impurity) were dissolved in 1dm³ of water. If 25cm³ of this solution were neutralized by 21cm³ of 0.1moldm³ HCL. Calculate the mass concentration of NaOH in gdm³.

Answer 1

NaOH + HCl ==> NaCl + H₂O

Converting volumes from cm³Â to dm³:

volume of HCl = 21.0 cm³Â = 21.0 ÷ 1000 = 0.021 dm³

volume of NaOH = 25.0 cm³Â = 25.0 ÷ 1000 = 0.025 dm³

Amount of HCl in mol = concentration in mol/dm³Â Ã— volume in dm³

= 0.1 mol/dm³Â Ã— 0.021 dm³

= 0.0021 mol

The equation shows that 1 mol of HCl reacts with 1 mol of NaOH. So the amount of NaOH in 25.0 cm³Â is 0.0021 mol.

[NaOH] = 0.0021 mol / 0.025 dm³ = 0.084 mol/dm³