Can you do questions 1 and 2, thanks! Q1) Calculate the concentration and uncert

Question

Can you do questions 1 and 2, thanks!

Q1) Calculate the concentration and uncertainty (absolute & %relative) of the standardized HCl solution. Q2) Calculate the grams of KOH determined to be in the potash along with the uncertainty (absolute & %relative) Q3) Calculate the percent purity of the potash sample along with the uncertainty (absolute %relative) MWkOH 65,1056 g/mol (negligible unc.) MWNa2Co3 105.98844 g/mol (negligible unc.) Standardization of HCI Na2co3 0.2105g (+0.0002 g) Volume used mLu 35.24 (+0.03 mL) Titration of potash sample vs HCI Volume used mLuc 38.48 (+0.03 mL) HCI HCI

Explanation / Answer

SOLUTION:

Q.1: Na2CO3(aq) + 2HCl(aq) -------- > H2CO3(aq) + 2NaCl(aq)

moles of Na2CO3 used = mass / molar mass = (0.2105 g +/- 0.0002 g) / 105.98844 g/mol

= (0.001986 +/- 2x10-6) mol

Let the concentration of HCl be 'M'

Volume of HCl used = (35.24 +/- 0.03) mL = (0.03524 +/- 0.00003) L

From the above balanced chemical reaction

2 x moles of Na2CO3 = moles of HCl

=> 2 x (0.001986 +/- 2x10-6) mol = M x (0.03524 +/- 0.00003) L

=> M = 2 x (0.001986 +/- 2x10-6) mol / (0.03524 +/- 0.00003) L

=> M = (0.1127 +/- 0.0002) mol/L (absolute uncertainity) (answer)

or M = (0.1127 +/- 0.1697 %) mol/L (relative uncertainity) (answer)

Q.2: moles of KOH = moles of HCl = M x V = (0.1127 +/- 0.0002) mol/L x (0.03848 +/- 0.00003) L

= (0.004337 +/- 9x10-6) mol

Hence mass of KOH = (0.004337 +/- 9x10-6) mol x 65.1056 g/mol = (0.2824 +/- 0.0006) g (answer)

or mass of KOH = (0.2824 +/- 0.1937%) (answer)

Q.3: Percent purity = [(0.2824 +/- 0.0006) / (0.5615 +/- 0.0002) g] x 100 = (50.29 +/- 0.1) % (answer)

or Percent purity = (50.29 +/- 0.2%) % (answer)

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