Question

Focus Question 1: Predict how many grams of KHP is needed to neutralize 5 mL of 1.0 M NaOH. Explain the basis for your prediction.

Answer 1

More KHP + NaOH -> kalap theo. ratio : Isole KP= I More Nooth Imole nlaon Neutsalize Imore Ktp. Litse = 1000m Naoh :- Molarity =10m Volume =5MI = 0.005L Mors =MXVE I OMX0.0052 = 5 x 10 moles I SX10 moles Nooth Neubalize 5x103 moins KHP. Mole & Weight Molecules waight Weight of kHpa moles of Ktp x Moleculas Weight of AHP = 5x108 moie A 204 12 % = 1.020

e KAP + NaOH t knap +H2O d i sole kHP = | Mole Naott More last Neutsalize Imole Ktp. Nach Molavity = M OHIM vomme EV = 5mi Tocomi Fizitze] 0.005LT moles a $1 XV = 0.180.0052 = 0.0005 moly. = 5x10 Moles. 5x10" Moies Naot Nentalize 5X10 Mons KHP- More = Weight More Moleculas weight Weight of kHp = nors X Molecula Weight of os # F of SHPKHD 5xio" por x 20112 = 0.1027: Naot Neutsali ze ktp = 0.102 gram's mor

e KAP + NaOH t knap +H2O d i sole kHP = | Mole Naott More last Neutsalize Imole Ktp. Nach Molavity = M OHIM vomme EV = 5mi Tocomi Fizitze] 0.005LT moles a $1 XV = 0.180.0052 = 0.0005 moly. = 5x10 Moles. 5x10" Moies Naot Nentalize 5X10 Mons KHP- More = Weight More Moleculas weight Weight of kHp = nors X Molecula Weight of os # F of SHPKHD 5xio" por x 20112 = 0.1027: Naot Neutsali ze ktp = 0.102 gram's mor

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e KAP + NaOH t knap +H2O d i sole kHP = | Mole Naott More last Neutsalize Imole Ktp. Nach Molavity = M OHIM vomme EV = 5mi Tocomi Fizitze] 0.005LT moles a $1 XV = 0.180.0052 = 0.0005 moly. = 5x10 Moles. 5x10" Moies Naot Nentalize 5X10 Mons KHP- More = Weight More Moleculas weight Weight of kHp = nors X Molecula Weight of os # F of SHPKHD 5xio" por x 20112 = 0.1027: Naot Neutsali ze ktp = 0.102 gram's mor