(1) How many moles of NaOH reacted in Trial 1? (2) What was the experimental mol

Question

(1) How many moles of NaOH reacted in Trial 1?

(2) What was the experimental molarity (M) for NaOH in Trial 1?

The NaOH used in this lab was 0.1M

Run 1
0.1388M C6H8O7   
ml       pH Derivative

0.00   12.96   -0.15  
.50      12.90   -0.20  
1.0      12.81   -0.43  
1.5      12.66   -1.10  
2.0      12.31   -2.84  
2.5      10.03   -4.81  
3.0       6.23   -4.74  
3.25   5.87   -2.16  
3.50   5.61   -0.97  
4.0       5.45   -0.60  
4.25   5.20   -0.64  
4.50   5.08   -0.55  
5.00   4.80   -0.52  
5.25   4.69   -0.47  
5.50   4.58   -0.44   

Mass of Beaker (step 1; g) 10.7g Mass of Beaker Citric Acid (step 3; g) 12.7g Mass of Citric Acid (step 1 step 3; g) 2g Mass of beaker citric acid after cooling (step 7; g) 12.8g Molarity of Citric Acid Solution (M) 0.1338 mols/L Molari Calculation Moles Masssubstance molar mass substance 2 192.12 0.0104 M Molarity Moles volume (L) 0.0104 0.075 0.1388 moles/L

Explanation / Answer

3 NaOH (aq) + H3C6H5O7 (aq) Na3C6H5O7 (aq) + 3 H2O(l)

H3C6H5O7(aq) + H2O(l)  <=> H2C6H7O7- + H3O+     Ka1 = 7.4x10-3
               H2C6H5O7- + H2O(l) <=> HC6H6O72- + H3O+    Ka2 = 1.7x10-5
               HC6H5O72- + H2O(l) <=> C6H5O73- + H3O+       Ka3 = 4.0x10-7

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