Help Needed: Balancing pH of Acetic Acid & NaOH Mixture

[Pengwuino]

I have a question here and i don't really know how to do it.

I need to know : What is the pH of a mixture of 30ml 0.1M acetic acid and 30ml of 0.1M NaOH? $$\[HC_2 H_3 O_3 = 1.8*10^{ - 5} \]$$

I'm not even sure what the equilibrium equation would be! Any help would be appreciated.

 

[siddharth]

Pengwuino, notice you have an acid (acetic acid) and a base (NaOH) .

• Initially, how many moles of each are there?
• When they react, what product will they form?
• How many moles of acid and base react, and how many moles of product is formed?
• So after, the reaction what happens to the product?

Figure out the answers to the above questions, and you are half way to the required answer.

 

[Blueshift5]

I would be happy to help you with balancing the pH of your acetic acid and NaOH mixture. First, it is important to understand that acetic acid (HC2H3O2) is a weak acid, while NaOH is a strong base. This means that when they are mixed, they will undergo a neutralization reaction, producing water and a salt (sodium acetate). The equilibrium equation for this reaction is as follows:

HC2H3O2 + NaOH → H2O + NaC2H3O2

To calculate the pH of the mixture, we need to first determine the moles of each substance present. This can be done using the formula: moles = concentration (in M) x volume (in L).

For acetic acid, we have 30 mL x 0.1 M = 0.003 moles.
For NaOH, we have 30 mL x 0.1 M = 0.003 moles.

Since the reaction is 1:1, we can see that both substances will be completely consumed in the reaction, leaving us with only the products (water and sodium acetate). This means that the final volume of the solution will be 60 mL.

To find the concentration of the products, we can use the formula: concentration (in M) = moles / volume (in L).

For water, we have 0.003 moles / 0.06 L = 0.05 M.
For sodium acetate, we have 0.003 moles / 0.06 L = 0.05 M.

Now, we can use the equilibrium constant (Ka) for the dissociation of acetic acid to calculate the pH of the solution. The equation for Ka is:

Ka = [H+][C2H3O2-] / [HC2H3O2]

We know the concentration of the products (water and sodium acetate) and the initial concentration of acetic acid (0.1 M). We can plug these values into the equation to solve for [H+].

Ka = [H+][C2H3O2-] / [HC2H3O2]

1.8 x 10^-5 = [H+][0.05] / [0.1]

[H+] = 1.8