a volume of 100 ml of a buffer of 1 m nh3 and 1m nh4^+ is placed in two half - cells connected by a salt bridge. a current of 1.5 a is passed through the cell for 20 minutes. if only the electrolysis of water takes place and the electrode reactions are;right electrode: 2h2o + o2 + 4e → 4 oh^- left electrode: 2h2o → 4h^+ + o2 + 4e, then, the ph of:
correct option is a) right electrode will increase
a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 we’re being asked to calculate the ph of 0.03m hc9h7o4.remember that weak acids partially dissociate in water and that acids donate h+ to the base (water in this case). the dissociation of hc9h7o4 is as follows:hc9h7o4 (aq) + h2o(l) ⇌ h3o+(aq) + c9h7o4– (aq)from this, we can construct an ice table. remember that liquids are ignored in the ice table and ka expression.the ka expression for hc9h7o4  is:ka=productsreactants=[h3o+][a-][hc9h7o4]note that each concentration is raised by the stoichiometric coefficient: [hc9h7o4], [h3o+] and [c9h7o4–] are raised to 1.plugin the value of ka and the terms from the ice chart: ka=[h3o+][a-][hc9h7o4]3.6 × 10-4=[x][x][0.03 -x]check if x is negligible in the denominator: if hc9h7o4ka>500 then x is negligible in the denominatorhc9h7o4ka= 0.03 3.6 ×10-4 = 83.3 < 500since the ratio is less than 500, we can’t remove (–x) from the equation. we first need to manipulate the equation to get a quadratic equation:3.6×10-4 = xx0.03-x0.03-x3.6×10-4   =x21.08 ×10-5-3.6×10-4 x = x2x2 +1.08 ×10-5-3.6×10-4 x =0we will then use the quadratic formula to solve for x. the quadratic formula is:x = -b±b2-4ac2afrom the equation, we have a = 1, b = 1.08×10–5, and c = –3.6 × 10–4. solving for x, the equilibrium concentration of [h3o+] and [c9h7o4–]:x = -b±b2-4ac2ax = -1.08×10-3±1.08×10-32-41-3.6 ×10-421x = -1.08×10-3±0.037962the calculation then splits into two ways:x = -1.08x10-3+0.037962x = 0.01844                          x = -1.08x10-3-0.037962x = -0.03904we take the positive value of x since the concentration of h3o+ and a– must increase. this means [h3o+] = [a–] = x = 0.01844 mnow calculate the ph using the formula below:ph = -log[h3o+]ph = -log[0.01844]ph = 1.73the ph of 0.030 m of aspirin is 1.73. we’re being asked to determine which of the given solutions is a good buffer system.recall that a buffer can be composed of the following:• weak base and its conjugate acid• weak acid and its conjugate base• weak acid and a strong base ▪ concentration of weak acid > concentration of strong base• weak base and a strong acid ▪ concentration of weak base > concentration of strong acid[readmore]a. 0.19 m potassium hydroxide + 0.27 m potassium bromide.potassium hydroxide → koh → group 1a k+ + oh- → strong basepotassium bromide → kbr breaks up in the solution: kbr → k+ + br-▪ k+ is a neutral ion → will not contribute to ph  hbr       +      h2o       →      br-      +         h3o+(weak acid)                            (conjugate base)•  strong base + weak base• not a good buffer systemb. 0.28 m hydroiodic acid + 0.18 m sodium iodide. hydroiodic acid → hi → weak acidsodium iodide → nai → breaks up in the solution: nai → na+ + i-hi       +      h2o       →      i-      +         h3o+(weak acid)                            (conjugate base)• weak acid + conjugate base = buffer• the solution is a good buffer systemc. 0.30 m ammonia + 0.37 m ammonium nitrate. ammonia → nh3 → weak baseammonium nitrate → nh4no3 → breaks up in the solution: nh4no3 → nh4+ + no3-nh3       +      h2o       →      nh4+      +         oh-(weak base)                            (conjugate acid)• weak base + conjugate acid = buffer• the solution is a good buffer systemd. 0.17 m hypochlorous acid + 0.12 m hydrobromic acid. hypochlorous acid → hclo → weak acidhydrobromic acid → hbr → weak acid•  weak acid + weak acid• not a good buffer systeme. 0.38 m calcium bromide + 0.30 m barium bromide.calcium bromide → cabr → breaks up in the solution: cabr → ca+ + br-barium bromide → babr → breaks up in the solution: babr → ba+ + br-•  conjugate base + conjugate base• not a good buffer systemanswer: the following aqueous solutions are good buffer systems b. 0.28 m hydroiodic acid + 0.18 m sodium iodide and c. 0.30 m ammonia + 0.37 m ammonium nitrate since the solution is a buffer, we can use the henderson-hasselbalch equation to calculate the ratio of [so32-]/[hso3-] to have a ph of 7.39:ph=pka+logconjugate baseweak acidin the solution, identify the weak acid and the conjugate base. recall from the bronsted-lowry definition that an acid is a proton (h+) donor and the base is a proton (h+) acceptor.• nahso3 is a salt that will give hso3- when it dissociatesnahso3 → na+ + hso3-• na2so3 is a salt that will give so32- when it dissociatesna2so3 → 2 na+ + so32-•hso3- is a weak acid.          - concentration of hso3- = concentration of nahso3• when hso3- loses h+ to give a conjugate base, so32- is formedhso32- - h+ → so32-- concentration of so32- = concentration of na2so3      hso3-    +   h2o     →        so32-        +       h3o+(weak acid)    (base)      (conjugate base)  (conjugate acid)we can rewrite the henderson-hasselbalch equation as:ph=pka+logso32-hso3-calculate [so32-]/[hso3-] ratio:• ph = 7.39• pka = 7.19ph=pka+logso32-hso3-7.39=7.19+logso32-hso3-logso32-hso3-=7.39-7.19logso32-hso3-log=0.2loglog has a base of 10 and when you divide a number by log you will take 10 to the power of the numberso32-hso3-=100.2[so32-]/[hso3-] = 1.58the ratio [so32-]/[hso3-] is 1.58 use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4 a) ans. concentration = no. of moles of solute * 1000/ vol. in m;

no. of moles of solute = concentration * vol / 1000 = 1.4 * 270 / 1000 = 0.378 = 0.378 mol.



b)

molality = 1.65

molality = moles of solute / mass of solvent in kg

1.65 = moles of solute / 5.5 x 10^-5 kg

moles of solute = 9.075 x 10^-5

c)

mass of sucrose = 75 x 1.4 / 100 = 1.05g

moles = 1.05 / 342 = 3.07 x 10^-3 from the figure:total resistance of the circuit, r = (rab + 15) ω = 20 ωcurrent in the circuit , \[i = \frac{v}{r} = \frac{8}{20} a\] ∴ voltage across ab, vab = i.rab = 2 vthe emf of the cell connected as above is given by: \[e = \frac{l}{l} v\_0\] here:l = 60 cm (balance point)ab = l = 1 m = 100 cm (total length of the wire) \[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 v\] for water, h2o1 mol h2o = 18.0 g h2o1 mole h2o = 6.022x1023 h2o molecules[readmore]the number of h2o molecules in a 9.00-g sample of water:9.00 g h2o×1 mol h2o18.0 g h2o×6.022×1023 h2o molecules1 mol h2o= 3.01x1023 h2o moleculesthe number of h2o molecules in a 9.00-g sample of water is 3.01x1023 h2o molecules. because of the small size of boron atom and presence of only six electrons in its valence shell, b(oh)3 accepts a pair of electrons from oh- ion of h2o, releasing a proton.the reaction mechanism is: 4 parts water to 1 part sugar--ex. 1 cup water and 1/4 cup sugar. 
mix together and just bring to a boil. 
remove from heat.
let cool to room temp.
use a feeder that has red on it--do not add coloring to the nectar.