Titration Lab Post-Lab Help?

Consider a 0.238 M aqueous solution of sodium hydroxide, NaOH.

a) How many grams of NaOH are dissolved in 23.46 mL?

b) How many individual hydroxide ions (OH−) are found in 23.46 mL?

c) How many moles of sulfuric acid, H2SO4, are neutralized by 23.46 mL of 0.238 M NaOH(aq)? [Hint: begin by writing a balanced equation for this neutralization reaction.]

1 Answer

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  • 3 weeks ago

    a)

    (0.238 mol NaOH/L) × (0.02346 L) × (39.99715 g NaOH/mol) = 0.223 g NaOH

    b) Supposing the NaOH dissociates completely:

    NaOH → Na{+} + OH{-}

    (0.238 mol NaOH/L) × (0.02346 L) × (1 mol OH{-} / 1 mol NaOH) ×

    (6.022 × 10^23 ions/mol) = 3.36 × 10^21 OH{-} ions

    c)

    H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

    (0.238 mol NaOH/L) × (0.02346 L) x (1 mol H2SO4 / 2 mol NaOH) =

    0.00279 mol H2SO4

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