Calculate the theoretical amount of .5M NaOH that it would take to neutralize approx. 2.0 g KHP. KHP is a monoprotic acid, the molar mass of KHP is 204.22 g/mol and the molar mass of NaOH is 40.00 g/mol. Answer in terms of mL.
I converted g KHP to mols, then mols of KHP to mols of NaOH. Then I converted mols of NaOH to g of NaOH and used the molarity equation (M=mols/liters) to get the volume of of NaOH. I got .783L or 783 mL is this correct?
3 answers
No. Your error is you convert mols NaOH to grams NaOH then used GRAMS for mols in M = mols/L but you can't do that. You should convert 2 g to mols KHP, then to mols NaOH, then mols NaOH/M = L NaOH
NaOH + KHPh => NaKPh + HOH
Need (2/204.22)mole NaOH = 0.0098-mole NaOH b/c Rxn is 1:1 Rxn Ratio. Ask what volume of 0.50M NaOH would deliver (2/204.22)mole of NaOH to neutralize 2g KHPh or 0.0098 mole KHPh? That is, (0.50M NaOH)(Volume of NaOH soln)= 0.0098 mole of NaOH delivered to neutralize 2 gms KHPh. Solve for Volume = 0.0196-Liter = 19.6 ml of the 0.5M NaOH(aq) solution will neutralize 2-gms KHPh (=0.0098 mole KHPh neutralized).
Need (2/204.22)mole NaOH = 0.0098-mole NaOH b/c Rxn is 1:1 Rxn Ratio. Ask what volume of 0.50M NaOH would deliver (2/204.22)mole of NaOH to neutralize 2g KHPh or 0.0098 mole KHPh? That is, (0.50M NaOH)(Volume of NaOH soln)= 0.0098 mole of NaOH delivered to neutralize 2 gms KHPh. Solve for Volume = 0.0196-Liter = 19.6 ml of the 0.5M NaOH(aq) solution will neutralize 2-gms KHPh (=0.0098 mole KHPh neutralized).
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