Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jennie28 on May 19, 2016, 10:30:19 AM
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A 25.0 sample of HCl is neutralize by 26.5mL of 0.155 M NaOH. Calculate the molarity of the HCl solution. (Hint: In this case, Molarity of HCl is MA in moles/liter of HCl.)
The equation would be M1V1/M2V2
(0.155M)(26.5mL)/(25.0mL)=.164M Is this right?
Do I multiply my answer by HCl moles which is 36.46g/mole?
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A 25.0 sample of HCl is neutralize by 26.5mL of 0.155 M NaOH. Calculate the molarity of the HCl solution. (Hint: In this case, Molarity of HCl is MA in moles/liter of HCl.)
What does it mean for this solution to be neutralised?
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To make the acid neutral
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The equation would be M1V1/M2V2
(0.155M)(26.5mL)/(25.0mL)=.164M Is this right?
Do I multiply my answer by HCl moles which is 36.46g/mole?
The first part seems sensible. What would the units be after you have multiplied M by molar mass? How would this differ, if for instance you had Ca(OH)2 and not NaOH?
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Would I divide .164 M/ 1 liter= .164 M/Liters of HCl?