40 mL of 0.30 weak acid, HA, (Ka = 4.2 X 10^-10), is titrated by 0.200M NaOH. calculate the pH after the following volumes of NaOH have been added.
(a) 0 mL
(b) 44.3 mL
(c) At the equivalence point
(d) 75.0 mL
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40 mL of 0.30 weak acid, HA, (Ka = 4.2 X 10^-10), is titrated by 0.200M...
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 46.19 mmol (millimoles) of HA and 2.24 mmol of the strong base. Then, more strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 85.1 mL ?
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. A solution is made by titrating 9.91 mmol (millimoles) of HA and 2.66 mmol of the strong base. Then, more strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 58.2 mL ?
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH.A.A solution is made by titrating 8.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH?Express the pH numerically to two decimal places.B.More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 43.0 mL?Express the pH numerically to two decimal places.
Titration of Weak Acid with Strong Base A certain weak acid, HA, with a Ka Value of 5.61 *10^-6, is titrated with NaOH. PART A A solution is made by mixing 8.00 mmol(millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH? express the pH numerically to two decimal places. pH = ? PART B More strong base is added until the equicalence point is reached. What is the pH of this solution at the...
50.00 mL of 0.100M of a weak acid (Ka=1.3x10-5) is titrated with 0.100M NaOH. a. Compute the volume of NaOH required to reach the equivalence point. b. Calculate the pH of the original solution before any NaOH has been added. c. After 30.00 mL of NaOH has been added, what is the pH of the solution? d. What is the pH at the equivalence point? e. Write a brief explanation as to why it is...
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?
36.53 mL of a 0.223 M solution of weak acid HA are titrated with 0.2 M NaOH. What is the pH of the solution after 8.25 mL of the NaOH have been added? The Ka for the weak acid is 0.0000077101.
33.79 mL of a 0.157 M solution of weak acid HA are titrated with 0.22 M NaOH. What is the pH of the solution after 12.98 mL of the NaOH have been added? The Ka for the weak acid is 0.0000005732.
3.)A certain weak acid, HA, with a Ka value of 5.61×10?6, is titrated with NaOH. Part A A solution is made by titrating 7.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. Part B More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 35.0 mL ?...
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?