Can I measure the molarity of Acetic Acid titrating Vineger with Sodium Hydroxide?

Imagine I have 5 g of vineger which I titrated with 33 mL of 0.1M Sodium Hydroxide (NaOH).

I tried V₁S₁=V₂S₂ formula but It needs the volume of the acid.

I have tried to calculate the volume with W=SVM/1000 formula which gives me 250/3S mL (where S is the molarity of Acetic Acid), but at the end S is cancelled out.

Is there any way I can calculate the molarity of Acetic Acid?

• You were able to weight 5 g in a container. Maybe you still have some of the vinegar, may record the weight of an empty graduated cylinder (as $m_0$). Then fill this cylinder again until it contains again 5 g (as $m_1 = m_0 + \pu{5 g}$) and read the volume of the vinegar at the wall of the graduated cylinder. The density should be somewhere between 1.00 and 1.30 g/cm³ (reference, acetic acid). – Buttonwood Feb 12 at 13:18
• According to the Handbook of Physics and Chemistry, the density of vinegar ($6$% $\ce{CH3COOH}$) is $1.0069 g/mL$ at $20°C$. So you can solve your problem yourself. – Maurice Feb 12 at 14:14
• @Maurice with the precision shown on the numeric values, I believe we can just write $1.0 g/mL$ – Nicolas Feb 12 at 14:43
• @Nicolas Yes we can ! – Maurice Feb 12 at 14:57
• Is there a good reason for deviating from standard variables' symbols for molarity $(c→S)$ and volume $(V→W)$? $S$ is entropy and $W$ is work, so one would have to bend their mind trying to read your formulas. Kinda reminds me of this episode of Andrew Dotson — Physics Professors Be Like (YouTube). – andselisk Feb 12 at 19:32

To use the formula that you indicate ($$V_1 * C_1 = V_2 * C_2$$), it is therefore necessary to know the volume of acetic acid $$V_1$$ or you only have the mass $$m_1$$. To switch from one to the other, you need to know / determine the density of the acetic acid used : you just need to take a volume $$V$$ of acetic acid and weigh it, the ratio $$\frac{m}{V}$$ corresponds to the density $$\rho_1$$ -> by transferring into the dosage equation you thus arrive at $$\frac {m_1}{\rho_1} C_1 = V_2 * C_2$$, you know everything except $$C_1$$ so you can determine the concentration
• Actually it should be $\frac {m_1} {\rho_1}$ to get volume. – MaxW Feb 12 at 22:46