1) 28.92mL of NaOH are required to titrate .5982 g of KHP to the phenolphthalein

Question

1) 28.92mL of NaOH are required to titrate .5982 g of KHP to the phenolphthalein endpoint. 22.41mL of the same NaOH solution are needed to titrate 25.00mL of the HCL solution to the phenolphthalein endpoint. If 34.62 mL of HCl solution are needed to titrate .5879g of the soda ash sample, what is the precent Na2CO3 in this sample?

2) Would failing to dry the poassium acid phthalate prior to its use cause the percent Na2CO3 in the soda ash sample to be greater than, less than, or equal to the true percent Na2CO3 in the soda ash sample? Why?

Explanation / Answer

(1) NaOH + KHP => NaKP + H2O

Moles of NaOH = moles of KHP = mass/molar mass of KHP

= 0.5982/204.22 = 0.0029292 mol


Molarity of NaOH = moles/volume of NaOH

= 0.0029292/0.02892 = 0.10129 M


HCl + NaOH => NaCl + H2O

Moles of HCl = moles of NaOH = volume x molarity of NaOH

= 22.41/1000 x 0.10129 = 0.0022698 mol


Molarity of HCl = moles/volume of HCl

= 0.0022698/0.02500 = 0.090793 mol


Na2CO3 + 2 HCl => 2 NaCl + CO2 + H2O

Moles of HCl = volume x molarity of HCl

= 34.62/1000 x 0.090793 = 0.0031433 mol


Moles of Na2CO3 = 1/2 x moles of HCl

= 1/2 x 0.0031433 = 0.0015716 mol


Mass of Na2CO3 = moles x molar mass of Na2CO3

= 0.0015716 x 105.99 = 0.16658 g


Mass% of Na2CO3 = mass of Na2CO3/mass of sample x 100%

= 0.16658/0.5879 x 100% = 28.33%


(2) If KHP is wet => it contains water

=> actual mass of KHP is lower

=> calculated NaOH molarity is lower

=> calculated HCl molarity is lower

=> calculated mass of Na2CO3 is lower

=> calculated mass% of Na2CO3 less than true value

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