(a)
Interpretation:
The common-ion effect needs to be explained.
Concept introduction:
Any atom or a molecule with a net charge is known as an ion. The net charge of an electron is opposite and equal to a proton. The net charge of an ion is never zero. This is due to the number of electrons being unequal to the number of protons.
Common ion effect is an effect in which several species of reversibly associated chemical solutions increase the concentration of any of its components by an equilibrium process.
(b)
Interpretation:
The use of a buffer solution to maintain a constant pH value needs to be explained.
Concept introduction:
A buffer solution is a mixture of a weak base and a conjugate acid or vice versa. The pH of a buffer solution changes very little when on adding a small amount of strong acid or a strong base to it.
The pH value indicates the acidic or basic nature of the given solution.
(c)
Interpretation:
The determination of pKa of a weak acid from a titration curve needs to be explained.
Concept introduction:
Titration is the process of identifying the concentrations of an unknown solution with the given concentrations of another solution.
The titration curve is used to plot the values determined after the process of titration.
It is an easy method of recording observations after the titration process.
pKa value determines the value of acidic strength. It is given by the negative log of an acid. This is the ka value.
Lower the pKa value stronger will be the acid.
(d)
Interpretation:
The measurement of pH with an acid-base indicator needs to be explained.
Concept introduction:
An ion or a molecule capable of donating a proton is known as an acid.
Substances that release hydrogen ions in an aqueous solution are called bases.
pH indicators are used to indicate the acidic or basic nature of a solution by changing its color.
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EBK GENERAL CHEMISTRY
- 11 (a) Define a buffer solution (b) What are the components of (i) an acidic buffer ? (ii) a basic buffer ?arrow_forwardA 250-mL buffer solution contains 0.0510 mole of KH2PO4 and 0.0875 mole of K2HPO4. (a) Calculate the molar concentrations of H2PO4–and HPO42–, respectively, in thesolution. (b) What is the pH of the solution? (H2PO4– has Ka = 6.2 x 10–8) (c) Write a net ionic equation for the buffering reaction against a strong acid by this buffer. (d) If 0.012 mole of hydrochloric acid (HCl) is added to the solution, create a reaction table to show the buffering reaction and calculate the molar concentration of H2PO4–and HPO42–, respectively, in the resulting solution after the buffering reaction. (e) What is the pH of the resulting solution?arrow_forwardA buffer solution contains 0.50 mol of hydrazoic acid (HN3) and 0.62 mol of sodium hydrazoate (NaN3) in 7.00 L. The K, of hydrazoic acid (HN3) is Ka = 1.9e-05.. (a) What is the pH of this buffer? pH = 5.96 (b) What is the pH of the buffer after the addition of 0.35 mol of NaOH? (assume no volume change) pH = 5.86 x pH = X (c) What is the pH of the original buffer after the addition of 0.35 mol of HI? (assume no volume change) xarrow_forward
- 7. You are given a 500 ml solution of 0.1 M HNO3 and an indicator Hin (Ka = 3.2x10) which changes from yellow (Hin) to green (In'). If you add two drops of indicator to the solution: (a) What colour is the solution initially? (b) What pH would the colour change become visible if you were adding 0.10M NaOH?arrow_forwardIf the pH of a buffer solution is equal to the pKa of the acidin the buffer, what does this tell you about the relative concentrationsof the acid and conjugate base forms of the buffercomponents?(a) The acid concentration must be zero. (b) The base concentrationmust be zero. (c) The acid and base concentrations must be equal. (d) The acid and base concentrations must be equal tothe Ka. (e) The base concentration must be 2.3 times as large asthe acid concentration.arrow_forwardThe pKa of a weak acid acid HA is 4.47. Consider the titration of 100 mL of 200 mM HA (aq.) with 200 mM NaOH (aq). Draw a graph representing this titration that includes the following labeled points/regions, which are deliberately out of order. (A) pH=7| (B) Starting point (0 mL NaOH added)' (C) Region where the pH approaches 13.30 (D) Equivalence point| (E) Buffered region" (F) Halfway point And determine the pH and mL NaOH added for points B, D, and Farrow_forward
- An aqueous buffer solution of volume 100 cm3 consists of 0.20 M CH3COOH and 0.20 M NaCH3CO2. pKa value of CH3COOH is 4.75. (a) Predict the pH value of this solution. 3 sig. fig. (b) Predict the pH value after adding 5.92 mmol NaOH (note, not mM. Volume is not 1 L) to this buffer solution assuming volume is not changed. 3 sig. fig.arrow_forwardChrome yellow, PbCrO4, is the yellow pigment used in road markings.At 15 °C, the solubility product of PbCrO4 is 1.69 × 10-14 mol2dm-6.(a) Write an expression for the solubility product of PbCrO4. [1](b) Calculate the solubility of PbCrO4 [3](c) On adding concentrated aqueous lead(II) nitrate dropwise to 0.010 mol dm-3 potassiumchromate (VI), what is the concentration of lead(II) ions when the first trace ofprecipitate appears? [3]arrow_forward- (a) Suppose you wanted to make a buffer of exactly pH 7.00 using KH,PO, and Na,HPO,. If the final solution was 0.1 M in KH,PO,, what concentration of Na,HPO, would you need? (b) Now assume you wish to make a buffer at the same pH, using the same substances, but want the total phosphate molarity ([HPO,²]+ [H;PO, ]) to equal 0.3. What concentrations of the KH,PO4 and Na;HPO, would be required?arrow_forward
- A buffer solution contains 0.35 mol of arsenous acid (H3ASO3) and 0.62 mol of sodium dihydrogen arsenite (NaH₂AsO 3) in 4.40 L. The K₂ of arsenous acid (H3ASO3) is K₂ = 5.1e-10. (a) What is the pH of this buffer? pH = (b) What is the pH of the buffer after the addition of 0.22 mol of NaOH? (assume no volume change) pH = (c) What is the pH of the original buffer after the addition of 0.18 mol of HI? (assume no volume change) PH =arrow_forwardA buffer is made by mixing 5.0 mL of 3.0 M NH,(aq) with 10 mL of 1.44 M NH Cl(aq). What is the pH of the buffer at 25°C, if the base ionization constant of ammonia is 1.8x10° and the ion product of water is 1.0x10-14 at this temperature?. Enter your answer in the box provided with correct units and sig. figs.: Comment : Henderson-Hasselbalch equations for acidic and basic buffers are listed on the next page. Answer: pH, 'bufferarrow_forwardAmmonia is a convenient buffer system in the slightly basic range. (a) What is the pH of a buffer solution containing 49 g of NH,CI dissolved in 1.00 L of 0.919 M NH3? pH = (b) How many moles of acid are required to change the pH of this solution by 0.05 pH units? mol (c) Suppose 4.6 mL of 12.6 M HCI solution is added to 471 ml of the solution of Part (a). Calculate the new pH. pH =arrow_forward
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