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# Answer to Question #26046 in Other Chemistry for Teenesha

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?
1
2013-03-13T09:40:07-0400
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&rsquo;s why NaOH is limiting reagent and amount of HCl needed for this reaction is 0.025 mol. It&rsquo;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

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