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Tx29 Tx29
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6 years ago
Calculate the molarity of Hydrochloric acid solution in the original 25mL sample..the molarity of NaOH is 0.1M.

Sample data table.

Molarity of base = 0.1M
Volume of acid = 25mL
Final buret reading = 35.7mL
Initial buret reading = 50.0mL
Volume of base used = 14.3mL

Please, show me step by step of how to solve this as I have more to complete.. Thanks in advance.

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wrote...
Staff Member
6 years ago
HCl + NaOH ----> NaCl + H2O

Molarity = mol/L
Volume of base (NaOH) = initial buret reading (50.0mL) - final buret reading (35.7mL) = 14.3 mL volume of base used to neutralize 25mL of HCl
 n = molarity of base (NaOH) x volume of base in L
 
Covert mL to L
1L = 1000mL

n = 0.1 M x 0.0143 L = 0.00143 mole of NaOH

Molarity of HCl = 0.00143 mole / 0.025 L= 0.0572 M of HCl

 
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