You have 2 moles of sodium hydroxide for every 1 mole of Tartaric acid After

You have 2 moles of sodium hydroxide for every 1 mole

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You have 2 moles of sodium hydroxide for every 1 mole of Tartaric acid. After completing the stoichiometry, we ended with a total of 9.35 x 10 -4 moles of tartaric acid. By taking the starting mass of Tartaric acid (.134g) and dividing it by this value, we got a molar mass of 143.32gH 2 Tart. We had a percent error of 4.51% since the actual molar mass of tartaric acid was 150.09g. This error could have derived from a couple of things. First, the acid could have been slightly impure when used in the titration overall affecting the amount of
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NaOH needed to be used therefore changing the value used in stoichiometry creating a different value to be calculated for moles of tartaric acid. Secondly, just as with the titration of NaOH and KHP, the error could have been subject due to the uneven amount of burette drops given out during titration as at one point we opened the nozzle a little more to speed up the process and then closed it to slow it down as we were trying to be as precise as possible. Conclusion Part 1 Claim: The molarity for the Sodium Hydroxide was found to be . 09640M. Evidence: In order to find the molarity for the Sodium Hydroxide (NaOH), a standardized solution if KHP was used for titration. The NaOH was titrated into the KHP until the equivalence point was reached which was determined by the color change from clear to light pink. Through stoichiometry, we were able to take the starting mass of .511g of KHP and convert it into moles using the molar mass of 204.23g KHP. We could then use a 1:1 mole ratio in order to convert to moles of OH - which gave us a total of . 00250mol OH - . We then divided this by the amount of NaOH used in titration to reach the equivalence point which was .02591L which gave us a molarity of .0964M NaOH. We did not run a second trial and therefore did not need to average out the totals of both trials so our total molarity for the sodium hydroxide was found to be .0964M. Reasoning (part 1 and 2): An acid –base titration is the determination of the concentration of an acid or base by exactly neutralizing the acid or base with an acid or base of known concentration to reach the equivalence point meaning that the amount of moles of the base is equal to the moles of the acid present (or when H + = OH - ) . This allows for quantitative analysis of the concentration of an unknown acid or base solution. When the titration occurs, a double replacement reaction happens between the acid and the base where water and salt are the products. Bases are proton acceptors where when they are dissolved in water, they produce
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OH - ions which are negatively charged. Reversely, acids are proton donors where when dissolved in water, they produce H + ions that are positively charged causing them to be attracted to the hydroxide ions produced by the base. Therefore, in a neutralization acid-base reaction, when the positively charged cations of hydrogen and the negatively charged anions of hydroxide are equivalent, they are bonded together to create the products of the reaction, water and salt.
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