[jennie28]

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?

[sjb]

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?

[jennie28]

To make the acid neutral

[sjb]

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?

[jennie28]

Would I divide .164 M/ 1 liter= .164 M/Liters of HCl?