The pH of a 4.0 × 10-3 mol dm-3 solution of weak base is 10.4. Calculate the Kb of the weak base.

Just b (i-ii)

A 25.0 cm³ sample of 0.620 mol dm-³ nitric acid was placed in a beaker and 38.2 cm³ of 0.550 mol dm-³ aqueous sodium hydroxide were added. Calculate the pH of the solution formed. Give your answer to 2 decimal places.

Suppose a group of students put 2.92 grams of an unknown solid acid into a 250-mL volumetric flask and filled it up to the mark with deionized water.  After this they pipetted out 15.75 mL of a 1.78 M NaOH(aq) solution into another 250-mL volumetric flask and filled it up to the mark with deionized water.  The students were given a pH meter that was already calibrated by their teaching assistant (T.A.) and they got the following result shown in the table below.  Calculate the Molarity of the unknown solid acid.  Hint:  You can approximate the endpoint of the titration by taking the average volume between two consecutive titrant volume reading and you can assume that it is a 1:1 stoichiometry ratio between the acid and OH-(aq).   VNaOH (mL) pH VNaOH (mL) pH 0 1.97 15 3.43 1 2.05 16 3.57 2 2.12 17 3.65 3 2.20 18 3.77 4 2.29 19 3.88 5 2.38 20 4.01 6 2.47 21 4.23 7 2.58 22 4.65 8 2.69 23…

Calculate the pH of a 10 M C5H5NHCl (aq) (pyridine chloride salt of pyridine) at 25 °C.  The base ionization constant of pyridine is Kb = 1.7×10–9.

What is the pH of a 0.10 M N(CH3)3(aq)? The base ionization constant  of trimethylamine is Kb= 6.5×10–5. Enter your answer with correct units and significant figures.

a) Calculate the pH of an aqueous solution of the salt NaA, which was made to a concentration of 0.080 M NaA.  The Ka of the weak acid HA is 2.0 x 10-5. Your pH should go to two places past the decimal point (i.e. 1.35). b) Aniline, C6H5NH2, has a pKb of 9.40.  What is the pH of a solution which is 0.12 M in aniline and 0.38 M in anilinium ion, C6H5NH3+? Your answer should be to two places past the decimal.

(b) Sodium hydroxide reacts with propanoic acid in the following equation: NaOH + CH3CH,COOHCH3CH2COONA + HLO A buffer solution is formed when sodium hydroxide is added to an excess of aqueous propanoic acid. (i) Calculate the number of moles of propanoic acid in 50.0 cm of 0.125 mol dm aqueous propanoic acid. (ii) Use your answers to part (b)(i) to calculate the number of moles of propanoic acid in the buffer solution when 2.00 cm' of 0.500 mol dm aqueous sodium -3 hydroxide are added to 50.0 cm' of 0.125 mol dm aqueous propanoic acid. ) Hence calculate the pH of this buffer solution at 298.15 K.

The pH of an aqueous solution of 0.400 M hydrocyanic acid, (Ka (HCN) = 4.00 × 10-10) is

Calculate the change in pH that results from adding 0.160 mol NaNO2 to 1.00 L of 0.160 M HNO2 (aq).  Ka HNO2 = 7.2 x 10-4