A solution of baking soda NaHCO_3 prepared from 0.755 g dissolved in 100 mL of water was used as...
Question:
A solution of baking soda {eq}\displaystyle \rm NaHCO_3 {/eq} prepared from {eq}\displaystyle \rm 0.755 \ g {/eq} dissolved in {eq}\displaystyle \rm 100 \ mL {/eq} of water was used as an antacid. {eq}\displaystyle \rm 25.00 \ mL {/eq} of {eq}\displaystyle \rm 0.960 \ M \ HCl {/eq}were used to digest the sample. The volume of {eq}\displaystyle \rm 0.595 \ M \ NaOH {/eq} used in the back titration was {eq}\displaystyle \rm 6.80 \ mL {/eq}. How many moles of {eq}\displaystyle \rm HCl {/eq} did the bicarbonate neutralize?
Back Titration:
Back titration refers to the type of titration where the solution containing the analyte is allowed to react with an excess amount of the reagent. The excess reagent is then titrated such that the excess amount allows the determination of the analyte concentration. Usually, back titration is used when the reaction is too slow, or when one of the reactants involved is volatile such as ammonia.
Answer and Explanation: 1
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View this answerIn order to determine the moles of HCl that the bicarbonate was able to neutralize, we need to calculate the amount of reacted HCl. This can be done...
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