Calculate the molarity of a barium hydroxide solution if you used 41.65 mL of it

Question

Calculate the molarity of a barium hydroxide solution if you used 41.65 mL of it to neutralize 1.190 g of potassium hydrogen phthalate. Ba(OH)_2(aq) + 2 KHC_8H_4 Write a balanced chemical equation for the reaction of hydrochloric acid with sodium hydroxide. Calculate the molarity of a hydrochloric acid solution if 34.21 mL of 0.0431 M sodium hydroxide neutralizes 25.00 mL of the acid solution. Write a balanced chemical equation for the reaction of phosphoric acid with sodium hydroxide. Calculate the molarity of a phosphoric acid solution if 34.21 mL of 0.0431 M sodium hydroxide neutralizes 25.00 mL of the acid solution.

Explanation / Answer

Part-1

find find the each moles using weight / molar mass formula

no of moles of potassium hydrogen phthalate = 1.19 g / 204.22 g/mol = 0.005827 mol

Ba(OH)2 is a diacidic base

so from balanced equation one mole of Ba(OH)2 required 2 mol of potassium hydrogen phthalate or

in other words one mole of pottasium hydrogen phthalate rquired 1/2 mol of Ba(OH)2

accordingly 0.005874 mol of potassium hydrogen phthalate required 0.00587 / 2 = 0.0029135 mol of BaOH required

now we have a moles of Ba(OH)2 and volume of Ba(OH)2

volume given = 41.65 mL convert in to liters = 0.04165 L

Molarity M = 0.0029135 mol / 0.04165 L

= 0.07 M

Part 2

Balanced equation is

NaOH + HCl ----> NaCl + H2O

Part -3

no of moles of NaOH = 0.0431 M x 0.03421 L = 0.0014744 mol

from the balanced equation it is clear that one mole of NaOH required one mole of HCl

so accordingly

0.0014744 mol of NaOH required 0.0014744 mol of HCl

now we have moles of HCL and volume of HCl

M = 0.0014744 mol / 0.025 L

= 0.05897 M

Part - 4

H3PO4 + 3NaOH ----> Na3PO4 + 3H2O

Part -5

no of moles of NaOH = 0.0431 M x 0.03421 L = 0.0014744 mol

from the balanced equation one mole of naOH required 1/3 mole of H3PO4

so moles of H3PO4 = 0.0014744 mol / 3 = 0.0004915 mol

Molarity = 0.0004915 mol / 0.025 L

= 0.01965 M

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