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What is the concentration of a sodium hydroxide (NaOH) solution if it takes 25 mL of 0.1 M hydrochloric acid (HCl) to neutralize 30 mL of the NaOH solution in a titration experiment?

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ago by (570 points)
To solve this problem, we can use the concept of stoichiometry and the equation:moles of acid = moles of baseSince the reaction between sodium hydroxide  NaOH  and hydrochloric acid  HCl  is a 1:1 reaction, we can write the equation as:moles of HCl = moles of NaOHWe are given the volume and concentration of HCl, so we can calculate the moles of HCl:moles of HCl = volume of HCl  in L   concentration of HCl  in mol/L moles of HCl = 0.025 L  0.1 mol/L = 0.0025 molNow we know the moles of NaOH, which is equal to the moles of HCl:moles of NaOH = 0.0025 molTo find the concentration of NaOH, we can use the formula:concentration of NaOH  in mol/L  = moles of NaOH / volume of NaOH  in L The volume of NaOH is given as 30 mL, which is equal to 0.03 L. Now we can calculate the concentration of NaOH:concentration of NaOH = 0.0025 mol / 0.03 L = 0.0833 mol/LTherefore, the concentration of the sodium hydroxide  NaOH  solution is approximately 0.0833 M.
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