50.0 mL of 0.090 M nitrous acid (HNO2, Ka = 7.1 x 10-4), is titrated with 0.100 M NaOH, requiring 45.0 mL of strong base to reach the equivalence point.
(a) What will be the pH after 35.0 mL of NaOH have been added?
(b) What will be the pH at the equivalence point?
(c) What will be the pH after 60.0 mL of NaOH have been added?
50.0 mL of 0.090 M nitrous acid (HNO2, Ka = 7.1 x 10-4), is titrated with...
2. A sample of 50 ml of nitrous acid (KA = 5.6 x 10“) is titrated with 0.070 M NaOH. The equivalence point is reached after the addition of 43.2 ml of the strong base. a) (3 marks) What is the pH of the solution at the equivalence point? b) (3 marks) What is the pH of the solution after 50 ml of NaOH has been added to the original nitrous acid solution?
2. A sample of 65 ml of nitrous acid (KA = 5.6 x 104) is titrated with 0.075 M NaOH. The equivalence point is reached after the addition of 44.5 ml of the strong base. a) (3 marks) What is the pH of the solution at the equivalence point? b) (3 marks) What is the pH of the solution after 60 ml of NaOH has been added to the original nitrous acid solution?
2. A sample of 60 ml of nitrous acid (KA= 5.6 x 104) is titrated with 0.075 M NaOH. The equivalence point is reached after the addition of 45.8 ml of the strong base. a) (3 marks) What is the pH of the solution at the equivalence point? I b) (3 marks) What is the pH of the solution after 60 ml of NaOH has been added to the original nitrous acid solution?
50.0 mL of 0.10 M HNO2 is being titrated with 0.20 M NaOH. What is the pH after 25.0 mL NaOH has been added? What is the pH after 35.0 mL NaOH has been added?
A 50.0-mL sample of 0.15 M butanoic acid, CH3CH2CH2COOH, is titrated with 0.30 M NaOH(aq). Ka for butanoic acid is 1.52 x 10-5 a) How many mL of NaOH(aq) are required to reach the equivalence point? b)What is the pH of the solution after 27.0 mL of NaOH(aq) have been added?
50.0 mL of 0.275 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point? Ka for HNO2 is 4.0x10^-4
A 15.0 mL sample of 0.21 M nitrous acid (HNO2), Ka = 4.6x10-4) is titrated with 0.15 M NaOH. Calculate the pH after the addition of the following volumes of base: a. 0.00 mL b. 10.00 mL c. 21.00 mL d. 25.00 mL
A student titrated a 100.0 mL sample of 0.100 M acetic acid with 0.050 M NaOH. (For acetic acid, Ka = 1.8 * 10^-5 at this temperature.) (a) Calculate the initial pH. (b) Calculate the pH after 50.0 mL of NaOH has been added. (c) Determine the volume of added base required to reach the equivalence point. (d) Determine the pH at the equivalence point?
A 50.0 mL sample of 0.25 M formic acid (HCOOH) aqueous solution is titrated with 0.125 M NaOH solution. Ka of HCOOH = 1.7 x 10−4. a. Calculate the pH of the solution after 50 mL of NaOH solution has been added. b. How many mL of 0.125 M NaOH need to be added to the sample to reach the equivalence point? What is the pH at the equivalence point?
A 40.0 mL sample of 0.150 M HNO2 (Ka = 4.60 x 10-4) is titrated with 0.200 M KOH. Calculate: a. the pH after adding 10.00 mL of KOH b. the pH at one-half the equivalence point c. the pH after adding 20.00 mL of KOH d. the volume required to reach the equivalence point e. the pH at the equivalence point f. the pH after adding 45.00 mL of KOH