Solve this chemistry practical for me

Chemistry

ScanSolve AI Answer

Answers

2 parts
b

0.101 M0.101 \text{ M}

c

3.69 g/dm33.69 \text{ g/dm}^3

Step-by-step solution

4 steps

Here is the solution to your chemistry practical problem.

a) To find the average volume of acid used, we identify the concordant readings (those within ±0.10 cm3\pm 0.10 \text{ cm}^3 of each other). In this case, all three readings (24.80, 24.70, 24.70) are concordant.

Step 1: Calculate the average volume of acid used.

Average volume of acid=24.80 cm3+24.70 cm3+24.70 cm33\text{Average volume of acid} = \frac{24.80 \text{ cm}^3 + 24.70 \text{ cm}^3 + 24.70 \text{ cm}^3}{3} Average volume of acid=74.20 cm33\text{Average volume of acid} = \frac{74.20 \text{ cm}^3}{3} Average volume of acid=24.73 cm3\text{Average volume of acid} = \boxed{24.73 \text{ cm}^3}

Part b· 1 step

To calculate the concentration of the acid, we use the titration formula. We assume the acid is monoprotic (e.g., HCl) and reacts with sodium hydroxide in a 1:1 molar ratio.

  1. Calculate the concentration of the acid.

    Given:
    Volume of base (VbV_b) = 25.0 cm325.0 \text{ cm}^3
    Concentration of base (CbC_b) = 0.10 M0.10 \text{ M}
    Average volume of acid (VaV_a) = 24.73 cm324.73 \text{ cm}^3 (from part a)

    Using the formula CaVa=CbVbC_a V_a = C_b V_b:

    Ca=CbVbVaC_a = \frac{C_b V_b}{V_a} Ca=0.10 M×25.0 cm324.73 cm3C_a = \frac{0.10 \text{ M} \times 25.0 \text{ cm}^3}{24.73 \text{ cm}^3} Ca=2.5024.73 MC_a = \frac{2.50}{24.73} \text{ M} Ca=0.101 MC_a = \boxed{0.101 \text{ M}}

Part c

3.69 g/dm33.69 \text{ g/dm}^3

To find the concentration of the acid in g/dm3^3, we multiply its molar concentration by its molar mass. We will assume the acid is hydrochloric acid (HCl) for this calculation.

  1. Calculate the concentration of the acid in g/dm3^3.

    Molar mass of HCl = 1.008 (H)+35.453 (Cl)=36.461 g/mol1.008 \text{ (H)} + 35.453 \text{ (Cl)} = 36.461 \text{ g/mol}
    Concentration in g/dm3^3 = Molar concentration ×\times Molar mass

    Concentration (g/dm3)=0.10109 mol/dm3×36.461 g/mol\text{Concentration (g/dm}^3) = 0.10109 \text{ mol/dm}^3 \times 36.461 \text{ g/mol} Concentration (g/dm3)=3.69 g/dm3\text{Concentration (g/dm}^3) = \boxed{3.69 \text{ g/dm}^3}

Part d· 1 step

For a strong acid-strong base titration, a suitable indicator is one that changes color around the neutral pH.

  1. Name a suitable indicator.

    A suitable indicator for this titration is phenolphthalein.

Part e· 1 step

The color of phenolphthalein at the endpoint of a strong acid-strong base titration is a very faint pink.

  1. State the color of the indicator at the end point.

    The color of phenolphthalein at the end point is faint pink.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Chemistry Questions

QSolve this chemistry questions for meQSolve this chemistry questionQSolve this chemistry questionsQAnswer for me this chemistry itemsQI need to understand all chemistry practical
See all Chemistry questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp