Plz help fill in the white empty boxes and answer the 4 questions would really a

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Plz help fill in the white empty boxes and answer the 4 questions would really appreicate it :)

Concentration of Hydrochloric Acid (M)0.0646 Temperature( 23 Titration 1 Titration 2 Titration 3 Volume of Saturated Calcium Hydroxide (mL) Initial Burette Reading (mL) Final Burette Reading (mL) Volume of Hydrochloric48 10 10 10 6 11.2 38.8 11.2 43.66 16.8 5.2 Acid (mL) Moles of Hvdrochloric Acid (moles) Concentration of Hydroxide M) Concentration of Calcium Hydroxide M) Concentration of Calcium Experimental Solubility of Calcium Hydroxide (g L1) Mean Experimental Solubility of Calcium Hydroxide (gL1) Standard Deviation (g L1) RSD (pph) Literature Solubility for calcium Hydroxide1 (gL1) //www.lime.org/documents/lime basics/lime-physical-chemical pdf

Explanation / Answer

1. (a): Titration-1: moles of Hydrochloric acid = MxV(L) = 0.0646 Mx4.8x10-3 L = 3.1008x10-4 mol

Titration-2: moles of Hydrochloric acid = MxV(L) = 0.0646 Mx5.2x10-3 L = 3.3592x10-4 mol

Titration-3: moles of Hydrochloric acid = MxV(L) = 0.0646 Mx5.6x10-3 L = 3.6176x10-4 mol

(b): Titration-1: concentration of hydroxide = moles of HCl / Vt(L) = 3.1008x10-4 mol / 14.8x10-3 L = 0.02095 M

Titration-2: concentration of hydroxide = moles of HCl / Vt(L) = 3.3592x10-4 mol / 15.2x10-3 L = 0.0221 M

Titration-3: concentration of hydroxide = moles of HCl / Vt(L) = 3.6176x10-4 mol  / 15.6x10-3 L = 0.02319 M

(c): Titration-1: concentration of calcium hydroxide = (concentration of hydroxide) / 2 = 0.010475 M

Titration-2: concentration of calcium hydroxide = (concentration of hydroxide) / 2 = 0.01105 M

Titration-3: concentration of calcium hydroxide = (concentration of hydroxide) / 2 = 0.011595 M

(d):Titration-1: [Ca2+] = concentration of calcium hydroxide = 0.010475 M

Titration-2: [Ca2+] = concentration of calcium hydroxide = 0.01105 M

Titration-3: [Ca2+] = concentration of calcium hydroxide = 0.011595 M

(e): Titration-1: solubility (g/L) = 0.010475 molx74.09 g/mol/L = 0.7761 g/L

Titration-2: solubility (g/L) = 0.01105 molx74.09 g/mol/L = 0.8187 g/L

Titration-3: solubility (g/L) = 0.011595 molx74.09 g/mol/L = 0.8591 g/L

(f): Mean experimental solubility of calcium hydroxide = (0.7761 + 0.8187 + 0.8591) / 3 = 0.8180 g/L

(g): standard deviation = 0.0415 g/L

(h) Relative standard deviation(RSD) = 5.07 %

We can calculate the values for the next table in a same manner

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