Answer to Question #26046 in Chemistry for Teenesha

Question

1 g of NaOH was added to 30 cm3 of 1 mol/dm3 HCl. How many cubic meters 0.1 mol/dm3 KOH solution will neutralize excess acid?

Answer 1

The chemical equation is

HCl + NaOH = NaCl + H₂O

n(NaOH) = m(NaOH)/M_W(NaOH) = 1/(23 +16 +1) = 0.025 mol
30 cm³ = 30 mL = 0.030 L
1mol/dm³ = 1 mol/L
n(HCl) = C x V = 1 x 0.030 = 0.030 mol

According to equation n(NaOH) = n(HCl), that’s why NaOH is limiting reagent and amount of HCl needed for this reaction is 0.025 mol. It’s 0.005 mol HCl is left. The chemical equation for the next neutralization reaction is

HCl + KOH = KCl + H₂O
n(KOH) = n(HCl) = 0.005 mol
V(L) = n(mol) / C(M) = 0.005 / 0.1 = 0.05 L = 5.0 x 10^-5 m³
Answer: V(L) = 5.0 x 10^-5 m³