Date Lab Sec. A. Molar Concentration of a Weak Acid Solution Sample no. Monoprot

Question

Date Lab Sec. A. Molar Concentration of a Weak Acid Solution Sample no. Monoprotic or diprotic acid? Trial I Trial 2 Molar concentration of NaOH (moVL) 2. Volume of weak acid (ml.) 3. I. 2S.O Buret readng of NaOH, initial (ml) 4. Buret reading NaOH at stoichiometric point, 32.0 30.0 final (mL) Standa class 32 5. Volume of NaOH dispensed (ml) 6 Instructor's approval of pH vs. VaoH graph 7. Moles of NaOH to stoichiometric point (mol) 8. Moles of acid (mol) 9. Molar concentration of acid (moUL) 10. Average molar concentration of acid (molUL)

Explanation / Answer

You need to determine the molar concentration of weak acid used for the titration. For this a simple equation can be applied. VAMA= VBMB where VA is the volume of acid , MA is the molarity of Acid. And VB is the volume of Base (NaOH). And MB is the molarity of Base ( NaOH). The molarity or molar concentration of NaOH is given as 0.09776 mol/L . And volume of Acid used for each titration ( VA) is given as 25 ml. VB is the volume of NaOH dispensed from the burette up to the stochiometric point (or End point) ( zero to End point) . From this data Molarity or molar concentration of Acid MA can be determined from the First equation since VAMA =VBMB Or VBMB÷VA. Consider the data from first titration (trial 1). Here volume of NaOH used is 32 ml. Applying this value into the equation gives (32×0.09776)/25= 0.12513 mol/L. For Trial 2 Molar concentration of Acid is (30×0.09776)/25 = 0.11731 mol/L. For trial 3 it is (32× 0.09776)/25 = 0.12513 mol/L. The average of this 3 value gives the average molar concentration of Acid. That is ( 0.12513+0.117312+0.12513 ) ÷ 3 = 0.122524 mol/L.

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