A 0.5035 g of potassium hydrogen phthalate (HC_8H_4O_4, a weak monoprotic acid)
ID: 530213 • Letter: A
Question
A 0.5035 g of potassium hydrogen phthalate (HC_8H_4O_4, a weak monoprotic acid) required 22.55 ml of a NaOH solution for titration to a phenolphthalein endpoint. What is the molarity of the NaOH solution? When 25.0 mL of 1.00 M HCI is mixed with exactly 25.0 ml of NaOH in a calorimeter, the temperature of the solution increased from 25.0 degree C to 27.55 degree C. If the specific heat of the solution is 4.18 J/(g middot degree C), what is delta H for the reaction in kJ/mol. Take the total volume of the solution as 50.0 mL of density of the solution as 1.00 g/mL.Explanation / Answer
5) Mass of Potassium Hydrogen pthalate = 0.5035g
Molar mass of Potassium Hydrogen otha late = 204.22g
No of mole of KOH = 0.5035g/204.22g= 0.002465
0.002465mol of KHP react with 0.002465mol of NaOH
Volume of NaOH = 22.55ml
Molarity of NaOH = (0.002465mol/22.55ml)×1000ml =0.1093M
6) HCl + NaOH -------> NaCl + H2O
Heat absorbed by Solution, q(s) = m × T × C(s)
Where, m = Mass of solution , 50gm
T = 27.55 - 25 = 2.55
C(s) = Specific hheat of solution , 4.184(J/ g )
q(s) = 50g × 2.55 × 4.184(J/ g)
= 533.46J
Heat absorbed by solution = Heat released by reaction
Therefore, H = -533.46J
This heat is for 0.025mol of H2O formed
Therefore, for 1mol of H2O , H = (533.46J/0.025mol) × 1mol = -21.34kJ/mol
Note: 1.Calorimeter specific heat is not mentioned in this problem
2. Molarity of NaOH not mentioned
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