3. (20 points) Consider the titration of 25.0 mL of 0.05 M Sna' with 0.100 M...
Consider the titration of 25.0 mL of 0.01000 M Sn2+ by 0.0500 M Tl3+ in 1 M HCl, using Pt and saturated calomel electrodes to find the endpoint. a.) Write the balanced titration reaction. b.) Write two different half-reactions for the indicator electrode. c.) Write two different Nernst equations for the cell voltage. d.) Calculate E at the following volumes of Tl3+: 1.00, 2.50, 4.90, 5.00, 5.10, and 10.0 mL. (This is problem 3 in Chapter 15 from Eigth Edition...
ans A C B how ? 14. Consider the titration of 25.00 mL of 0.0500 M Sn2+ with 0.100 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+, using Pt and saturated calomel electrodes SCE l Sn4+, Sn2+, Fe3+, Fe3 | Pt(s) The titration reaction: 2Fe3 + Sn2+Snt+ 2Fe2+ The two half-reactions for the indicator electrode: Fe3+e Sn 2e Sn2+ Indicate the two Nernst equations for the cell voltage. E 0.732 V E 0.139 V Fe2+ IFe2 log...
5. Consider the titration of 25.0 mL of 0.0100 M Sn2 by 0.0500 M Ti Pt and saturated calomel electrodes to find the end point. (total 15 pt) M HCI, using (c) Write two different Nernst equations for the cell voltage. (3 pt) (d) Calculate E at the following volumes of TI: 1.0, 2.5,4.9,5.0,5.1, and 10.0 mL. (6 pt) 5. Consider the titration of 25.0 mL of 0.0100 M Sn2 by 0.0500 M Ti Pt and saturated calomel electrodes to...
A 100.0 mL solution of 0.0400 M Fe2+ in 1 M HCIO s titrated with 0.100 M Ce4+ resulting in the formation of Fe3+ and Ce3+. A Pt indicator electrode and a saturated calomel electrode are used to monitor the titration. Write the balanced titration reaction titration reaction:> Complete the two half-reactions that occur at the Pt indicator electrode. Write the half-reactions as reductions half-reaction: Ce e half-reaction: Fe e- E 0.767 V Select the two equations that can be...
You are titrating 100.0 mL of 0.0400 M Fe2+ in 1 M HCIO4 with 0.100 M Ce4+ to give Fe3+ and Ce3+ using Pt and calomel electrodes to find the endpoint. (a) Write the balanced titration reaction. > (b) Complete the two half reactions for the Pt electrode. Ce té E° = 1.70 V Fe +e = E° = 0.767 V [Fe2+11 1 - 0.241 E= 0.767 – 0.05916) (c) From the list in the column at the right, select...
A 100.0 mL solution of 0.0400 M Fe2+ in 1 M HCIO, is titrated with 0.100 M Ce*+ resulting in the formation of Fe+ and Ce3+. APt indicator electrode and a saturated calomel electrode are used to monitor the titration Write the balanced titration reaction. titration reaction:-> Complete the two half reactions that occur at the Pt indicatorelecrode Write the half-reactions as reductions half-reaction: Ice + e-→ We were unable to transcribe this imageсез+] . 0.241 (Ce+] 「 0.241 E...
A 25.0 mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 ✕ 10-5) is titrated with 0.100 M KOH solution. Calculate the pH after the addition of the following amounts of KOH. 0.0 mL 4.0 mL 8.0 mL 12.5 mL 20.0 mL 24.0 mL 24.5 mL 24.9 mL 25.0 mL 25.1 mL 26.0 mL 28.0 mL 30.0 mL
Мар You are performing a titration of 25.0 mL of 0.0100 M Sn4 in 1 M HCI with 0.0500 M Ag* to give Sn2 anu 3+ Ags using a Pt indicator electrode and a saturated calomel electrode (SCE) as the reference electrode. A) Write the balanced titration reaction B) Complete the two half-reactions that occur at the indicator electrode (shown is their corresponding reduction potential) Sn E 0.139 V 3+ Ag® Eo 1.90 V C) Choose the two different expressions...
Titration of Strong acid with strong base 2. Consider the titration of 25.0 mL of 0.100 M HCI with 0.100 M NaOH. a) Write down the chemical equation. Hellmunt Hell Hotele lua b) Calculate the volume of NaOH required to reach the equivalence point. Rome 250ML 0.25 0. In 2008 was x I c) Calculate the initial pH of the acid solution. (before adding NaOH). pol of Helin 0.1ac plte -log [ol 1] d) Calculate the pH after adding 5.00...
Consider the titration of 100.0 mL of 0.100 M methylamine (CH3NH2) with 0.500 M HNO3. Calculate the pH at the following volumes of acid added. For CH3NH3+, pKa = 10.632 (a) Find the equivalence point volume. (b) 0 mL (c) 9.0 mL (d) 10.0 mL (e) 20.0 mL (f) 30.0 mL