molar concentration of a Weak acid solution Molar concentration of NaOH (mol/L)
ID: 503460 • Letter: M
Question
molar concentration of a Weak acid solution Molar concentration of NaOH (mol/L) 2. Volume of weak acid (mL) 3. Buret reading of NaOH, Initial(mL) 4. Buret reading NaOH at stoichiometric point, final(mL) 5. Volume of NaOH dispensed(mL) 6. Instructor's approval of pH vs, V_ NaOH graph 7. Moles of NaOH to stoichiometric point (mol) 8. Moles of acid (mol) 9. Molar concentration of acid (mol/L) 10. Average molar mass of add (mol/L) Molar mass and the PK_a of a solid weak acid sample no.__ Monoprotic or diprotic acid? ____suggested mass____1. Mass of dry, solid acid(g) 2. Molar concentration of NaOH (mol/L) 3. Buret reading of NaOH, initial(mL) 4. Buret reading NaOH at stoichiometric point, final(mL) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of PH versus V_NaOH graph 7. Moles of NaOH to stoichiometric point(mol) 8.Moles of acid(mol) 9. Molar mass of acid (g/mol) 10. Average molar mass of acid(g/mol) 11. Volume of NaOH halfway to stoichiometric point(mL) 12. PK_a1 of weak acid(from graph) 13. Average PK_a1 14. How calculations for trial 1 on the next page.Explanation / Answer
A. Molar concentration of weak acid solution
1. NaOH molarity 0.1607 M
2. Weak acid (ml) 40.12 ml
5. NaOH (ml) used 11.08 ml
7. NaOH (moles) 0.1607 M x 11.08 ml = 1.78 mmol
8. weak acid (moles) 1.78 mmol
9. molarity weak acid moles/volume = 1.78 mmol/40.12 ml = 0.044 mol/L
Similarly other trial calculations can be done and from it average molarity of the weak acid solution can be calculated
B. Molar mass of solid weak acid
1. mass of acid 0.803 g
2. molarity NaOH 0.1607 M
5. Vol. NaOH used 30.94 ml
7. moles NaOH 0.1607 M x 30.94 ml = 4.972 mmol
8. moles of acid 4.972 mmol
9. molar mass acid grams/moles = 0.803 g/4.972 x 10^-3 mol = 161.50 g/mol
This calculation was done for Trial 1. Similarly other trials values can be used and calculations can be done. From it average molar mass can be calculated.
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