Exploring Substances: Acidic, Basic, and Neutral – Numerical Problems with Stepwise Solutions

Q1. A list contains 12 common household liquids. Out of these, 5 are acidic and 4 are basic. How many liquids are neutral?

Given: Total liquids = 12, acidic = 5, basic = 4, neutral = ?

Stepwise Solution:

1. Total liquids = acidic + basic + neutral
2. Let number of neutral liquids be x.
3. So, 12 = 5 + 4 + x
4. 12 = 9 + x
5. x = 12 − 9 = 3

Answer: There are 3 neutral liquids in the list.

Q2. In a class, students bring 18 food samples to test. If 7 samples are acidic and 5 are neutral, how many are basic?

Given: Total = 18, acidic = 7, neutral = 5, basic = ?

Stepwise Solution:

1. Total samples = acidic + basic + neutral
2. 18 = 7 + basic + 5
3. Add acidic and neutral: 7 + 5 = 12
4. So, 18 = 12 + basic
5. basic = 18 − 12 = 6

Answer: There are 6 basic samples.

Q3. A supermarket sells 40 bottles of different liquids in a day. 16 bottles are acidic drinks, 10 bottles are basic cleaning liquids and the rest are neutral products. Find the number of neutral products sold.

Given: Total bottles = 40, acidic = 16, basic = 10, neutral = ?

Stepwise Solution:

1. Total = acidic + basic + neutral
2. 40 = 16 + 10 + neutral
3. 16 + 10 = 26
4. So, 40 = 26 + neutral
5. neutral = 40 − 26 = 14

Answer: 14 neutral products were sold.

Q4. Out of 25 items in a kitchen, 9 are sour (acidic), 8 are bitter/soapy (basic) and the rest are neutral. What fraction of the items are neutral?

Given: Total = 25, acidic = 9, basic = 8, neutral = ? (fraction)

Stepwise Solution:

1. Neutral items = Total − (acidic + basic)
2. Neutral items = 25 − (9 + 8) = 25 − 17 = 8
3. Fraction of neutral items = Neutral ÷ Total = 8 ÷ 25

Answer: The fraction of neutral items is 8/25.

Q5. In a science fair, a group displays 15 labelled solutions: the ratio of acidic : basic : neutral solutions is 3 : 2 : 1. Find the number of acidic solutions.

Given: Total = 15, ratio = 3 : 2 : 1 (acidic : basic : neutral)

Stepwise Solution:

1. Total parts in ratio = 3 + 2 + 1 = 6 parts
2. One part = Total ÷ 6 = 15 ÷ 6 = 2.5
3. Acidic solutions = 3 parts = 3 × 2.5 = 7.5
4. But number of solutions must be a whole number. So, this ratio is only approximate and we round to the nearest whole number.
5. Approximate acidic solutions ≈ 8 (basic ≈ 5, neutral ≈ 2, total = 15)

Answer: Approximately 8 solutions are acidic (based on the given ratio).

Q6. A student tests 10 solutions with blue litmus paper. Blue litmus turns red in 4 solutions and does not change in the others. How many of the 10 solutions are not acidic?

Given: Total = 10, acidic (blue → red) = 4, not acidic = ?

Stepwise Solution:

1. Not acidic solutions = Total − acidic
2. Not acidic = 10 − 4 = 6

Answer: 6 solutions are not acidic (they may be basic or neutral).

Q7. In a lab, 18 test tubes are used to check turmeric indicator. Turmeric turns reddish-brown in 7 test tubes (showing bases). In the remaining test tubes, turmeric stays yellow. How many test tubes do not contain a base?

Given: Total = 18, basic = 7, non-basic = ?

Stepwise Solution:

1. Non-basic test tubes = Total − basic
2. Non-basic = 18 − 7 = 11

Answer: 11 test tubes do not contain a base.

Q8. A science teacher prepares 5 samples (A, B, C, D, E) and tests them with china rose indicator.

• 2 samples turn the indicator red.
• 2 samples turn it green.
• 1 sample shows almost no change.

How many samples are likely acidic, basic and neutral?

Given: Total = 5, red = acidic, green = basic, no change = neutral

Stepwise Solution:

1. Acidic samples = 2 (turn indicator red)
2. Basic samples = 2 (turn indicator green)
3. Neutral samples = 1 (no noticeable change)

Answer: 2 acidic, 2 basic and 1 neutral sample.

Q9. Out of 30 solutions tested in a lab, 12 change litmus colour and 18 do not change litmus at all. What percentage of solutions are neutral?

Assumption: Solutions that do not change litmus are taken as neutral for this question.

Given: Total = 30, neutral = 18, % neutral = ?

Stepwise Solution:

1. Percentage of neutral solutions = (neutral ÷ total) × 100
2. = (18 ÷ 30) × 100
3. First simplify 18 ÷ 30: 18/30 = 3/5 = 0.6
4. 0.6 × 100 = 60

Answer: 60% of the solutions are neutral.

Q10. A student wants to test 9 liquids using litmus paper. She has 12 strips of red litmus and 9 strips of blue litmus. If she uses 1 red and 1 blue strip for each liquid, how many red and blue strips will be left unused?

Given: Liquids = 9, red strips = 12, blue strips = 9

Stepwise Solution:

1. Red strips used = 1 × 9 = 9
2. Blue strips used = 1 × 9 = 9
3. Red strips left = 12 − 9 = 3
4. Blue strips left = 9 − 9 = 0

Answer: 3 red strips and 0 blue strips will be left.

Q11. A doctor advises Riya to take 2 antacid tablets a day for 5 days to neutralise extra acid in her stomach. How many antacid tablets will she take in total?

Given: Tablets per day = 2, days = 5

Stepwise Solution:

1. Total tablets = Tablets per day × Number of days
2. Total tablets = 2 × 5 = 10

Answer: She will take 10 antacid tablets in total.

Q12. A small antacid tablet can neutralise 3 “units” of acid in the stomach. If a person has 15 units of extra acid, how many such tablets are needed to neutralise it completely?

Given: 1 tablet → 3 units acid, Total acid = 15 units, tablets = ?

Stepwise Solution:

1. Number of tablets = Total acid units ÷ units neutralised by 1 tablet
2. = 15 ÷ 3
3. = 5

Answer: 5 tablets are needed to neutralise 15 units of acid.

Q13. A farmer adds 20 kg of lime powder to neutralise acidic soil in one small field. For a bigger field he needs 3 times this amount. How much lime powder does he need for the bigger field?

Given: Lime for small field = 20 kg, for bigger field = 3 times small field

Stepwise Solution:

1. Lime needed for bigger field = 3 × 20 kg
2. = 60 kg

Answer: He needs 60 kg of lime powder for the bigger field.

Q14. In a simple experiment, 10 mL of an acid is exactly neutralised by 10 mL of a base. If a student has 30 mL of the same acid, how many mL of this base will be needed to neutralise it?

Given: 10 mL acid ↔ 10 mL base; for 30 mL acid, base = ?

Stepwise Solution:

1. For equal strength, equal volumes neutralise each other.
2. Factor increase in acid volume = 30 ÷ 10 = 3
3. So base volume must also be 3 times: 10 × 3 = 30 mL

Answer: 30 mL of the base will be needed.

Q15. A gardener uses 4 packets of a mild basic powder to treat acidic soil in one month. In the next month, the soil becomes more acidic and he uses 6 packets. By how many packets did his use of basic powder increase?

Given: First month = 4 packets, second month = 6 packets, increase = ?

Stepwise Solution:

1. Increase in packets = Second month − First month
2. = 6 − 4
3. = 2 packets

Answer: His use of basic powder increased by 2 packets.

Q16. In a certain city, it rains on 20 days in a month. On 8 of those days, the rainwater is found to be acidic due to pollution. On how many days is the rainwater not acidic?

Given: Total rainy days = 20, acidic rain days = 8, non-acidic days = ?

Stepwise Solution:

1. Non-acidic rainy days = Total rainy days − acidic rainy days
2. = 20 − 8
3. = 12 days

Answer: Rainwater is not acidic on 12 days.

Q17. A lake originally has 500 fish. After a few years of acid rain, the number of fish reduces to 350. By how many fish has the population decreased? What percentage decrease does this represent (approximate to the nearest whole number)?

Given: Original = 500, new = 350

Stepwise Solution:

1. Decrease in number = Original − New
2. = 500 − 350 = 150 fish
3. Percentage decrease = (Decrease ÷ Original) × 100
4. = (150 ÷ 500) × 100
5. 150 ÷ 500 = 0.3
6. 0.3 × 100 = 30%

Answer: The fish population decreased by 150 fish, which is a 30% decrease.

Q18. A hill station receives 60 cm of rainfall in a season. Due to increased air pollution, scientists estimate that 40 cm of this rainfall is now acidic. What fraction and percentage of the rainfall is acidic?

Given: Total rainfall = 60 cm, acidic = 40 cm

Stepwise Solution:

1. Fraction of acidic rain = Acidic ÷ Total = 40 ÷ 60
2. 40 ÷ 60 = 4 ÷ 6 = 2 ÷ 3
3. Percentage acidic = (2 ÷ 3) × 100 ≈ 66.67%
4. Rounded to nearest whole number: ≈ 67%

Answer: About 2/3 or approximately 67% of the rainfall is acidic.

Q19. Four solutions have pH values 2, 5, 7 and 9. (a) Which solution is the most acidic?
(b) What is the difference in pH between the most acidic solution and the neutral solution?

Given: pH values = 2, 5, 7, 9; neutral pH = 7

Stepwise Solution:

1. (a) On the pH scale, smaller values (below 7) are more acidic.
2. The smallest pH here is 2, so pH 2 solution is most acidic.
3. (b) Difference in pH between pH 2 and pH 7 = 7 − 2 = 5 units.

Answer: (a) The solution with pH 2 is most acidic. (b) The pH difference between it and the neutral solution is 5 units.

Q20. Two solutions X and Y have pH values 4 and 9 respectively. How many pH units away from neutral (pH 7) is each solution? Which one is acidic and which one is basic?

Given: pH of X = 4, pH of Y = 9, neutral pH = 7

Stepwise Solution:

1. Distance of X from neutral = |7 − 4| = 3 units
2. Distance of Y from neutral = |9 − 7| = 2 units
3. pH < 7 means acidic, so X (pH 4) is acidic.
4. pH > 7 means basic, so Y (pH 9) is basic.

Answer: X is 3 pH units away and is acidic; Y is 2 pH units away and is basic.