Question

Part 1.)

Calculate the pH of each of the following strong acid solutions.

(a) 0.00555 M HClO₄

pH =

(b) 0.314 g of HBrO₄ in 21.0 L of solution

pH =

(c) 39.0 mL of 3.50 M HClO₄ diluted to 1.90 L

pH =

(d) a mixture formed by adding 59.0 mL of 0.00582 M HClO₄ to 16.0 mL of 0.00676 M HBrO₄

pH =

Part 2.)

Using values from Appendix C of your textbook, calculate the value of K_eq at 298 K for each of the following reactions:

(a) Fe₂O₃(s) + 3 CO(g) 2 Fe(s) + 3 CO₂(g)
K_eq =  

(b) 2 NO(g) N₂(g) + O₂(g)
K_eq =  

(c) 2 NOCl(g) 2 NO(g) + Cl₂(g)
K_eq =  

Part 3.)

(a) In the electrolysis of aqueous NaNO₂, how many liters of NO(g) (at STP) are generated by a current of 87.0 A for a period of 61.4 min? The unbalanced chemical reaction representing this electrolysis is shown below. NaNO₂(aq) + H₂O(l) NO(g) + O₂(g) + NaOH(aq)
__ liters of NO(g) is generated by this electrolysis.

(b) How many moles of NaOH(aq) are formed in the solution in this process?
__ moles of NaOH(aq) are formed.

Answer 1

Parts.) ph of solution is defined as negative logrthum of hydrogen im concentration. - pH = -log (ht) a) 0.00555 M Hdo4 H404 Copi) -> H+ (0%) + 604 (al.) [H+)= 0.00555M DH= -log (0.00555) = b) 0.314g of HBrog in 21.02 mol of HBrO4= given mass molar mass

Mol of HBYO4 = 0.3147 -0.00217 mol 144.61% molt Molarity = mola HBrO4 volume of solution 0.00217 mol = 0.0001034m 210L HBT04 = 0.0001034M concentrahon pH = -log [ht] s-log (0.00010 34) | 54 = 3.98 39.0m2 of 3.50M HU04 diluted to 1.902 Finding concentration of hloq in diluted solution (undiluted) Cavo = Calz (diluted) 3:50MX 3.90ML = (2x1.902 3:50M X 39,0m2 = -2 1.90L 3.50M+ 0.0394 =62 1.904 0.0718 M = (2 Concentrating H004 = 0.0718 m Dua-log 10.0718) 24 - 1.14 soluim = 1.14

HU04 saomL H BY04 16.0mL 0.00676 0.00SEUM HU04 ) - +004(a) C 0.00512M H B804 (1) H+ () + B804 (ap) 0.00676M Finding molgut mol y 4+ produced by H404 = = (im L0.01) Molarity x volume 0.005 1 mol X 0.01g: 0, 00031 24 n| mol 74+ produced by HØY04 = 0.00676 mol X 0.0162 = 0.0001 mol Total mol = 0.000 34 34 mol + 0.coull mol = 0.0004534 moll Total volume = 0.059L+ 0.0162 = 0.0751 concentratim of ut = 0.00045 34 mol - - 0.0060s M 0.075 L ph= -log[he] = -log (0.00605] [PA - a་ ༢ ། buj Bol = 2.2g Ansines