0 B. Molar Concentration of an Acid Solution Acid type, HA or H A: HCl Unknown N

Question

0 B. Molar Concentration of an Acid Solution Acid type, HA or H A: HCl Unknown No. Balanced equation for neutralization of acid with NaOH. Po Sample 3 Sample 1 Sample 2 25.0 1. Volume of acid solution (mL) 2. Buret reading of NaOH, initial (mL.) 3. Buret reading of NaOH. final (mL) 4. Volume of NaOH dispensed (mL) 5. Molar concentration of NaOH (mol/L), Part A 6. Moles of NaOH dispensed (mol) 7. Molar concentration of acid solution (mol/L) 8. Average molar concentration of acid solution (mol/L) 9. Standard deviation of molar concentration 25.0 25.0 0 20.1 Data Analysis, B Data Analysis, Data Analysis, D 10. Relative standard deviation of molar concentration (%RSD)

Explanation / Answer

Molar concentration NaOH you have take from Part A you didn't give that i will assume a value and do the calculation you just replace it with the molarity you have and follow same procedure

Molar concentration of NaOH = 1 mol/L (or M) = M1

Volume of NaOH V1 = 20.3 20.7 20.1 ml

= 0.0203 0.0207 0.0201 L

Moles of NaOH n1 = V1 * M1 = 0.0203 0.0207 0.0201 moles

Let volume of solution of acid V2 = 20 ml

V1 * M1 = V2 * M2

Molar concentration of acid be M2 = V1 * M1 / V2

M2 = 1.015 1.035 1.005 mol/L

average molar concentration of acid = (1.015 + 1.035 + 1.005) / 3 = 1.01833 mol/L

standard deviation of molar concentraion = 0.0153

% RSD = 1.5 %

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