Question #26046
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?
Expert's answer
The chemical equation is
HCl + NaOH = NaCl + H2O
n(NaOH) = m(NaOH)/MW(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
HCl + NaOH = NaCl + H2O
n(NaOH) = m(NaOH)/MW(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
