A solution of 15.05mL of NaOH is mixed with 45.00mLof 0.03102 M H 2 SO 4 (standa

Question

A solution of 15.05mL of NaOH is mixed with 45.00mLof 0.03102 M H2SO4 (standardized) and allowedto react. The solution (H2 SO4 andNaOH) is still acidic, so it is titrated with 11.85mLofstandardized KOH (0.09001 M) until the equivalence point isreached. What was the original concentration (molarity) ofthe NaOH solution?

Explanation / Answer

2NaOH + H2SO4 ->Na2SO4 + 2H2O 2KOH + H2SO4 ->K2SO4 + 2H2O c(KOH) = n/v n(KOH) = cv = 0.09001 x 0.01185 = 0.0010666mol n(H2SO4) remaining after reacted with NaOH =1/2 x n(KOH) = 0.000048mol Original n(H2SO4) = cv = 0.03102 x 0.045 =0.0013959mol n(H2SO4) reacted with NaOH = 0.001359 -0.000048 = 0.0013479mol n(NaOH) = 2/1 x n(H2SO4) = 0.0026958mol c(NaOH) = n/v = 0.0026958/0.01505 = 0.179M Hope this helps!

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