A) Write balanced molecular and net ionic equation for the neutralization of NAO

Question

A) Write balanced molecular and net ionic equation for the neutralization of NAO H by KHO (Mol Wt =204.23). KHp is an abbreviation for potassium hydrogen phthalate, KHC8H4O4, A substance of known high purity with one acid hydrogen as indicated by the formula.
B) if 2.08 g of KHP is dissolved in water and titrated with 40.00 mL of NaOH to the phenolphthalein endpoint, what is the molarity of the NaOH solution?
C) this is now standardize solution of NaOH is used to find the molecular weight of an acid H2A which has two acid hydrogens write a balanced molecular equation for this titration reaction.
D) if 1.08 g of this acid requires 37.18 mL of NaOH solution for the titration to the appropriate endpoint, calculate the molecular weight of the unknown acid, H2A. A) Write balanced molecular and net ionic equation for the neutralization of NAO H by KHO (Mol Wt =204.23). KHp is an abbreviation for potassium hydrogen phthalate, KHC8H4O4, A substance of known high purity with one acid hydrogen as indicated by the formula.
B) if 2.08 g of KHP is dissolved in water and titrated with 40.00 mL of NaOH to the phenolphthalein endpoint, what is the molarity of the NaOH solution?
C) this is now standardize solution of NaOH is used to find the molecular weight of an acid H2A which has two acid hydrogens write a balanced molecular equation for this titration reaction.
D) if 1.08 g of this acid requires 37.18 mL of NaOH solution for the titration to the appropriate endpoint, calculate the molecular weight of the unknown acid, H2A. A) Write balanced molecular and net ionic equation for the neutralization of NAO H by KHO (Mol Wt =204.23). KHp is an abbreviation for potassium hydrogen phthalate, KHC8H4O4, A substance of known high purity with one acid hydrogen as indicated by the formula.
B) if 2.08 g of KHP is dissolved in water and titrated with 40.00 mL of NaOH to the phenolphthalein endpoint, what is the molarity of the NaOH solution?
C) this is now standardize solution of NaOH is used to find the molecular weight of an acid H2A which has two acid hydrogens write a balanced molecular equation for this titration reaction.
D) if 1.08 g of this acid requires 37.18 mL of NaOH solution for the titration to the appropriate endpoint, calculate the molecular weight of the unknown acid, H2A.

Explanation / Answer

A) the balanced molecular equation is given by

NaOH + KHC8H404 ---> KNaC8H404   + H20

the ionic equation is given as

Na+ + OH- + K+ + HC8H404- ---> K+ + Na+ C8H4042- + H20

the net ionic equation is given by

HC8H404- + OH- ----> C8H4042- + H20

B)

we know that

moles = mass / molar mass

given

mass of KHP = 2.08 g

mola rmass of KHP = 204.23

so

moles of KHP = 2.08 / 204.23 = 0.0101846

the reaction is given by

NaOH + KHC8H404 ---> KNaC8H404   + H20

from the above reaction we get

moles of NaOH required = moles of KHP

so

moles of NaOH required = 0.0101846

now

molarity = moles / volume (L)

so

molarity of NaOH = 0.0101846 / 40 x 10-3

molarity of NaOH = 0.2546

so

the molarity of NaOH solution is 0.2546

C) given a diprotic acid

the reaction is given by

H2A + 2NaOH --->   Na2A + 2H20

D)

moles = molairty x volume (L)

so

moles of NaOH = 0.2546 x 37.18 x 10-3

moles of NaOH = 9.466 x 10-3

from the above reaction we get that

moles of H2A = 0.5 x moles of NaOH

so

moles of H2A = 0.5 x 9.466 x 10-3

moles of H2A = 4.733 x 10-3

now

moles = mass / molar mass

given

mass of acid H2A = 1.08 g

so

4.733 x 10-3 = 1.08 / molar mass

molar mass = 228.18 g

so

the molecular weight of the acid H2A is 228.18 g/mol

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.