What would be the pH of a solution obtained by mixing 5 g of acetic acid?

What would be the pH of a solution obtained by mixing 5 g of acetic acid?

pH=4. 49.

What would be the of a solution obtained by mixing 5g of acetic acid and 7.5 g of sodium acetate and making the volume equal to 500 mL?

Mol CH3COOH = 5g/(60g/mol) = 0.833 mol of CH3COOH/500mL or (1000mL/500mL) X 0.833 mol of CH3COOH/L solution = 1.67 mol of CH3COOH/L solution. Mole of CH3COONa = 7.5g/(82g/mol) = 0.091 mol of CH3COONa/500mL or (1000mL/500mL) X 0.091 mol of CH3COONa/L solution = 0.182 mol of CH3COONa solution.

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What is the pH of acetic acid and sodium acetate?

Next, solid sodium acetate is added to the acetic acid solution until the color of the indicator in the solution is “green” corresponding to pH = 7. Both the water and the acetic acid/acetate solution have the same color and therefore both solutions have same pH.

What is the pH of sodium acetate?

7
Sodium acetate has a pH of 7.

Why is the pH of acetic acid low?

Acetic acid is transported across the membrane from the feed phase in its unionized molecular form and not in its ionic form that is acetate. This required that the pH of the feed phase is lower than the pKa value of acetic acid which is 4.76 (at 25 degC).

How do you calculate the pH of a solution of acid?

Example #1: Calculate the pH of a solution obtained by mixing 100. mL of an acid of pH = 3.00 and 400. mL solution of pH = 1.00. 1) Determine total moles of H + in solution: 0.040 mol + 0.00010 mol = 0.0401 mol Note: we ignore the H + contribution from water. 2) Determine the [H +] of the combined solution: 3) Determine the new pH:

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How do you calculate N(acetic acid) to N(NaOH)?

n(acetic acid) = 10 × 0.1 = 1 mmol n(acetic acid) = n(NaOH) = 1 mmol This amount of NaOH is present in 10 mL of solution. Then you should calculate the pH of this solution (which contains a weak base, sodium acetate) to check your assumption about the indicator.

What is the equation for acetic acid with Ka = ka?

Acetic acid is a weak acid with Ka = 1.86 × 10. –5 and in this case c. weak acid >>> Ka, that is the equation to use is: [H+] =. Ka ⋅c weak acid = (1.86 10 ) 0.1. −5 × × = 0.0013638 M pH = -log[H+] = -log(0.0013638) = 2.865.

What is the pH of a solution obtained by mixing NaOH and ch3coh?

The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH is with 100 ml of 0.2 M NaOH would be: [Note : pKa for CH3COOH = 4.74 and log 2 = 0.301) ]. C OOH = 4.74 and log2 = 0.301) ]. Was this answer helpful?

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