Olutions that contain a weak acid, ha, and its conjugate base, a−, are called buffer solutions because they resist drastic changes in ph. when a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. for buffers containing a weak acid, the principal reaction is ha(aq)+h2o(l)⇌h3o+(aq)+a−(aq) unbuffered, a weak acid would ionize and the net reaction would proceed forward to reach equilibrium. however, when a significant amount of conjugate base is already present, the extent of the net reaction forward is diminished. thus, equilibrium concentrations of ha and a− are approximately equal to their initial concentrations. visualizing buffers you can visualize a buffer solution containing approximately equal concentrations of ha and a− in action using the following equation, where ka is the equilibrium constant: ka[h3o+]==[h3o+][a−][ha]ka[ha][a−] when the concentrations of ha and a− are relatively large, the addition of a small amount of acid or base changes their concentration negligibly. the result is that the ratio [ha]/[a−] maintains a value approximately equal to 1, and the [h3o+] is approximately equal to the ka value under buffer conditions. part a calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for hobr is 2.0×10−9. express your answer numerically

pH = 8.477

Explanation:

KOBr → K+  +  OBr-

∴ ni    0.3       -          -

  nf      -        0.3       0.3

HOBr  ↔ H3O+  +    OBr-

∴ ni   0.50          -              0.3            

  nf  0.5 - X        X           0.3 + X

∴ Ka = 2.0 E-9 = ([H3O+]*[OBr-]) / [HOBr]

⇒ Ka = 2.0 E-9 = ((X/1L)*(0.3 + X)/1 L) / ((0.5 - X)/1L)

⇒ 2.0 E-9 = ( 0.3X + X² ) / (0.5 - X)

⇒ X² + 0.3X - 1 E-9 = 0

⇒ X = 4.333 E-9 M

according Henderson-Hauselbach:

pH = pk + Log [A-] / [HA]

∴ [OBr-] = 0.3 mol/ 1 L + 4.333 E-9 M = 0.300 M

∴ [HOBr] = 0.5 mol / 1 L - 4.33 E-9 M = 0.500 M

∴ pKa = - log Ka = - Log ( 2.0 E-9 ) = 8.6989

⇒ pH = 8.6989 + Log ( 0.300/0.500 )

⇒ pH = 8.477

The answer to your question really depends on three things: how much water you have, how cold it is to start out, and how cold the things around it are. water actually freezes when it gets to 32 degrees fahrenheit (0 degrees celsius), but the time it takes to get there may be different.  let's start with the first. if you take two glasses, and fill one with a tiny bit of water, and the other about halfway, then put them both in the freezer, the one with less water will freeze first (you can try this at home, but i recommend using plastic cups and not glass ones).  now let's move on to the second part. let's say you have two glasses, and you fill one with really cold water that has been in the refrigerator, and the other with really hot water from the sink. if you put both of them in the freezer, the one that started out colder will freeze first.  for the third part, let's imagine that you have two glasses with the same amount of water in them, and the water is at the same temperature. imagine putting one outside on a really really cold day in georgia, and having a friend in alaska put one outside on the same day. since it would be so much colder in alaska, the glass of water there would freeze before yours.  so, if you took a tiny bit of really cold water in a glass, and put it outside on a cold day in alaska, it would freeze a lot faster than a big glass of hot water outside on a cold day in georgia.