question archive If you wanted to determine how much KOH KOH (potassium hydroxide; a strong base) was dissolved in water, you could titrate it with a known amount of KHP KHP (potassium hydrogen phthalate; a weak acid) until a color change indicates that the base is completely neutralized
If you wanted to determine how much KOH KOH (potassium hydroxide; a strong base) was dissolved in water, you could titrate it with a known amount of KHP KHP (potassium hydrogen phthalate; a weak acid) until a color change indicates that the base is completely neutralized
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If you wanted to determine how much KOH
KOH (potassium hydroxide; a strong base) was dissolved in water, you could titrate it with a known amount of KHP
KHP (potassium hydrogen phthalate; a weak acid) until a color change indicates that the base is completely neutralized. The molecular equation below depicts this neutralization reaction.
KHP(aq)+KOH(aq)→K
2
P(aq)+H
2
O(l)
KHP(aq)+KOH(aq)→K2P(aq)+H2O(l)
Write the net ionic equation that depicts the neutralization of the base KOH
KOH by KHP
KHP. Instead of using the entire chemical formula of phthalate, use HP
−
HP− to represent the acidic hydrogen phthalate ion and P
2−
P2− to represent the phthalate ion.
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The Molecular Neutralisation Reaction of Potassium Hydrogen Phthalate ion and Sodium Hydroxide is :
KHC8H4O4 (aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l)
Or
KHP (aq) + NaOH (aq) → KNaP(aq) + H2O(l)
The net ionic equation is:
HC8H4O4-(aq) + OH-(aq) → C8H4O42-(aq) + H2O(l)
Or
HP-(aq) + OH-(aq)→ P2-(aq) + H2O(l)
Please see the attached file for the complete solution
