Hello, i don\'t just need the answer I really want to learn how to do this. plea

Question

Hello, i don't just need the answer I really want to learn how to do this. please label and draw arrows? thank you very much

Calculate the molarity of a Ba(oh)2 solution if a 37.25 ml portion of the solution is neutralized by 45.00ml of 1.500 M HCl.

In standardizing a potassium hydroxide solution, a student finds that 32.60 mL if the base solution is needed to titrate 0.5045 g of KHP to a phenolphthalein end point. In another titration, 26.63 mL of this same base solution is needed to titrate 0.2003 grams of an unknown solid monoprotic acid.

a.)    Determine the molarity of the base solution

b.)    Determine the molar mass of the unknown solid monoprotic acid.

Explanation / Answer

(1) Ba(OH)2 + 2 HCl => BaCl2 + 2 H2O

Moles of HCl = volume x concentration of HCl

= 45.00/1000 x 1.500 = 0.0675 mol


Moles of Ba(OH)2 = 1/2 x moles of HCl

= 1/2 x 0.0675 = 0.03375 mol


Molarity = moles/volume of Ba(OH)2

= 0.03375/0.03725

= 0.9060 M


(2) (a) KOH + KHP => K2P + H2O

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

= 0.5045/204.22 = 0.0024704 mol


Molarity = moles/volume of KOH

= 0.0024703/0.03260

= 0.07578 M


(b) Let the acid be HA

HA + KOH => KA + H2O


Moles of HA = moles of KOH = volume x concentration of KOH

= 26.63/1000 x 0.07578 = 0.002018 mol


Molar mass of HA = mass/moles of HA

= 0.2003/0.002018

= 99.26 g/mol

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