Fall 2016 periment 20 relab questions: No Prelab Questions Report sheet Postlab)
ID: 1064103 • Letter: F
Question
Fall 2016 periment 20 relab questions: No Prelab Questions Report sheet Postlab) questions solve on this page 1 Calculate the molarity (M) of NaoH solution which required 22.15 mL of NaoH to react completely with 0.552 g KHP. This problem is similar to what is in the Lab and calculation steps shown (Part A).. how all steps by using conversion factors. Must finish with c.v. and fv. 2 Calculate the volume (mL) of 0.450 M of KoH required to neutralize 25.55 mL the of 0.155 M HCl solution. Show and rearranged Equation before solving get problem. Show all steps with conversion factor and formula, equations etc.to full credit.Explanation / Answer
The reaction between KHP and NaOH is
KHC8H4O4(aq) + NaOH(aq) --> KNaC8H4O4(aq) + H2O(l)
Atomic weights : C=12, H=1, K=39, O=16
Molar mass of KHP= 39+1+12*8+4+4*16=204
Moles of KHP in 0.552gm = mass/molar mass
Moles of KHP =0.552/204=0.002706
As per the reaction, 1 mole of KHP requires 1 mole of NaOH
0.002706 moles require 0.002706 moles of NaOH
Volume of NaOH= 22.15ml
Volume in L= Volume in ml/1000 = 22.15/1000
Molarity = moles/ Volume in (L)= 0.002706*1000/22.15=0.1203M
2.
The reaction between KOH and HCl is KOH+HCl---àKCl+H2O
Atomic weights : K=39, H=1, Cl=35.5, O=16
Molar masses : KOH= 39+1+16=56, HCl= 36.5
Moles of HCl in 22.55ml of 0.155M = Molarity* Volume(L)= 0.155*22.55/1000=0.003495
Moles of KOH as per the reaction =same as moles of HCl =0.003495
Volume= Moles/Molarity =0.003495/0.45 =0.0078 L
3. since the concentration of KHP will be less due to presence of NaCl as contaminant, the volume of KHP in the sample will be less and hence moles of NaOH required will be less. Since molarity= moles/unit volume. Molarity of NaOH will be higher.
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