I need help with filling in the rest of this. Confused on what steps to take to

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I need help with filling in the rest of this. Confused on what steps to take to calculate this eb B. Molar Concentration of an Acid Solution Acid type: HA Unknown No. Balanced equation for neutralization of acid with NaOH. Sample 1 Sample 2 Sample 3 25.00 1. Volume of acid solution (mL) 2. Buret reading of NaOH, initial (mL) 3. Buret reading of NaOH, final (mL) 25.00 25.00 IT rn 65mL Volume of NaOH dispensed (nb) Molar concentration of NaOH (mol L), Part A agan yr0203 mol/L Moles of NaOH dispensed (mol) Molar concentration of acid solution (mon) ) Average molar concentration of acid solution (mol L) 9. Standard deviation of molar concentration Appendix B da Relative standard deviation of molar concentration (%RSD) Appendix B

Explanation / Answer

Ans. # Balanced Reaction:   HA(aq) + NaOH(aq) -------> NaA(aq) + H2O(l)

According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol acid.

Trial 1: Volume of NaOH consumed = 28.4 mL = 0.0284 L

#6. Moles of NaOH dispensed = Molarity x Volume of NaOH dispensed in liters

                                                = 0.303 M x 0.0284 L

                                                = 0.0086052 mol

#7. According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol acid. So, moles of acid in 25.0 mL sample must be equal to the moles of NaOH dispensed.

Hence, moles of acid = 0.0086052 mol

Now,

            Molarity of HA = Moles of HA/ Volume of solution in liters

                                                = 0.0086052 mol / 0.025 L

                                                = 0.344208 M

Trial 2: Volume of NaOH consumed = 25.90 mL = 0.0259 L

#6. Moles of NaOH dispensed = 0.303 M x 0.0259 L = 0.0078477 mol

#7. According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol acid. So, moles of acid in 25.0 mL sample must be equal to the moles of NaOH dispensed.

Hence, moles of acid = 0.0078477 mol

Now,

            Molarity of HA = 0.0078477 mol / 0.025 L = 0.313908 M

Trial 3: Volume of NaOH consumed = 25.60 mL = 0.0256 L

#6. Moles of NaOH dispensed = 0.303 M x 0.0259 L = 0.0077568 mol

#7. According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol acid. So, moles of acid in 25.0 mL sample must be equal to the moles of NaOH dispensed.

Hence, moles of acid = 0.0077568 mol

Now,

            Molarity of HA = 0.0077568 mol / 0.025 L = 0.310272 M

#8. Average molarity = (0.344208 + 0.313908 + 0.310272) M/ 3 = 0.322796 M

#9. SD [AH]= 0.0186

#10. % RSD = 5.77 %

Note: SD and %RSD is calculated using https://www.miniwebtool.com/relative-standard-deviation-calculator/

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