Henderson- Hasselbalch equation is basically used for the calculation of pH during salt formation after adding base on the acid. Part a and part b are given below.
Here in part b the concentration of the salt = (0.3*25)/35 where 35 is total concentration (25+10).
In part c when 30 ml NaOH is added then concentration of excess HO - will be (5*0.3)/40 = 0.0375. so POH= -log(0.0375) = 1.4259. so the corresponding pH will be 14-1.4259 = 12.5741
A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH...
A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added base. K of CH3CH2COOH is 1.3 x 10-5. base Henderson-Hasselbalch equation: pH = pK+ log NaOH CH3CH2COOH Parta): 1) After adding 18.0 mL of the NaOH solution, the mixture is (Select) equivalence point on the titration curve. 2) The...
A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added base.Ka of CH3CH2COOH is 1.3 × 10−5.Henderson–Hasselbalch equation: p H = p K a + log [ b a s e ] [ a c i d ]Part a):1) After adding 18.0 mL of the NaOH solution, the mixture is ...
A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 18.0 mL of 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture. Ka of CH3CH2COOH is 1.3 \times × 10−5. Henderson–Hasselbalch equation: LaTeX: pH=pK_a+\log\frac{\left[base\right]}{\left[acid\right]} p H = p K a + log [ b a s e ] [ a c i d ] 2046 HW10 Q1-1.jpg 1) After adding the NaOH...
Question 6 1 pts A 10.0 mL sample of 0.25 M NH3(aq) is titrated with 0.20 M HCl(aq) (adding HCl to NH3). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added acid. Ko of NH3 is 1.8 x 10-5 Henderson-Hasselbalch equation: pH = pka + log og HCI NH, NH3- Parta): 1) After adding 10 mL of the HCl solution, the mixture is (Select] the equivalence...
1)A 10.0 mL sample of 0.25 M NH3(aq) is titrated with 0.20 M HCl(aq) (adding HCl to NH3). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added acid.Kb of NH3 is 1.8 × 10−5.Henderson–Hasselbalch equation:Part a):1) After adding 10 mL of the HCl solution, the mixture is [ Select ] ["at", "before", "after"] the equivalence point on the titration curve.2) The pH of the solution after...
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 15.0 mL of 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture. Assume that the volumes of the solutions are additive. 1)After adding the HCl solution, the mixture is [select one](before, After, at) the equivalence point on the titration curve. 2)The pH of the solution after adding HCl is [select one](7.00,1.40,11.00,12.60).
assignment will be closed on Friday, March 29th at 12.00 PM (noon). The correct answers will be available on Friday March 29th. at 12:05 PM Question 1 1 pts A 10.0 mL sample of 0.25 M NHsloq) is titrated with 10.0 mL of 0.20 M HCloa) (adding HCI to NHal. Determine which region on the titration curve the mixture produced is in, and the pH of the mixture. Kb of NH3 is 1.8 × 10-5. Henderson-Hasselbalch equation: pH base] HCI...
can u help? im not sure about my answers Question 6 2 pts A 50.0 mL sample of 0.20 M HCl(aq) is titrated with 0.10 M NaOH(aq) (adding NaOH to HCI). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added base. Assume that the volumes of the solutions are additive. NaOH HCI Parta) [Select ] after 1) After adding 20 mL of the NaOH solution, the...
A 20 mL sample of 0.01 M propionic acid (CH3CH2COOH; Pka = 4.87) is titrated with 0.05 M NaOH. A) Write out the chemical reaction for this titration. B) Calculate the initial pH of the sample. C) Calculate the volume of NaOH required to reach the equivalence point. D) Calculate the pH of the solution at the equivalence point. E) Sketch a titration curve for this titration (pH versus volume NaOH added). Note the location of the equivalence point on...
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?