Asked by Michelle
0.688g of antacid was treated with 50.00ml of 0.200M HCl. The excess acid required 4.21 mL of 1.200 M NaOH for back-titration. What is the neutralizing power of this antacid expressed as mmol of HCl per gram of antacid?
I think the answer should be:
(50.00ml HCl)(0.200M HCl)= 10mmol HCl
10mmol HCL/0.688g antacid= 14.535mmol/g.
I don't know what needs to done with 4.21ml NaOH?
I think the answer should be:
(50.00ml HCl)(0.200M HCl)= 10mmol HCl
10mmol HCL/0.688g antacid= 14.535mmol/g.
I don't know what needs to done with 4.21ml NaOH?
Answers
Answered by
DrBob222
Caroline has a problem just above yours. Here is a link.
http://www.jiskha.com/display.cgi?id=1428332431
http://www.jiskha.com/display.cgi?id=1428332431
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