I think you posted only part of the problem. I assume you have titrated the NaOH with HCl and you want to know the M of the NaOH.
mols HCl = M x L = ?
mols NaOH = mols HCl from the molar ratio of 1:1.
M NaoH = mols NaOH/L NaOH. You know mols NaOH and L NaOH, solve for M NaOH.
If 30 ml of 02 M HCL with 45 ml what is the molarity of NaOH
2 answers
I assumed that 45 mL of NaOH, and that "02 M" is 0.02 M HCl.
Write the balanced equation:
HCl + NaOH ---> NaCl + H2O
Then get the moles of HCl by multiply the volume by the molarity:
M = n/V
n = V*M
n = 30 mL * 0.02 M
n = 0.6 mmol HCl
Since in the reaction it's 1 mol HCl reacted for every 1 mol NaOH present, then the number of moles of NaOH is
0.6 mmol HCl * (1 mmol NaOH / 1 mmol HCl) = 0.6 mmol NaOH
Finally, solve for the molarity of NaOH.
M = n/V
M = 0.6 mmol / 45 mL
M = ?
Units in mol/L. Hope this helps~ `u`
Write the balanced equation:
HCl + NaOH ---> NaCl + H2O
Then get the moles of HCl by multiply the volume by the molarity:
M = n/V
n = V*M
n = 30 mL * 0.02 M
n = 0.6 mmol HCl
Since in the reaction it's 1 mol HCl reacted for every 1 mol NaOH present, then the number of moles of NaOH is
0.6 mmol HCl * (1 mmol NaOH / 1 mmol HCl) = 0.6 mmol NaOH
Finally, solve for the molarity of NaOH.
M = n/V
M = 0.6 mmol / 45 mL
M = ?
Units in mol/L. Hope this helps~ `u`