Chemistry Question?

Question

how do you do these questions? my final is in 3 days, and I'm unsure:

what's the pH?

1) 25.0 ml of 0.10 M HCl are mixed with 30.0 ml of 0.075 M NaOH

2) 10.0 ml of 0.10 M HCl, 5.0 ml of 0.19 M H2SO4 and 0.060g of NaOH

please help thanks!

Answer 1

The reaction

HCl + NaOH ----> NaCl + OH- + H+

shows that one mole of either HCl or NaOH will produce one mole of H+. When the compounds are mixed, equal moles of each will produce equal moles of H+ and OH-, which is neutral. If there are not equal moles, the excess compound will determine the pH. You are adding 25*0.1 millimoles of HCl to 30*0.075 millimoles of NaOH. Multiply these out to get

2.5 millimoles HCl
2.25 millimoles of NaOH

all but 0.25 millimoles of HCl gets reacted (neutralized). The moles of H+ in 0.25 millimoles of HCl is 0.25 millmoles. This total is in 25+30 = 55 ml of solution, so the molar concentration of H+ is 0.25/55 = 0.0045 M. The pOH is -log(0.0045) = 2.35.

In the second, you have 1 millimole of HCl, 0.95 millimole of H2SO4; this latter will produce 2*0.95 or 1.9 millimole of H+, which when added to the HCl contribution gives a total of 2.9 millimole of H+. 0.06 g of NaOH = 0.06/40 = .0015 mole of NaOH which provides 0.0015 mole or 1.5 millimole of OH-, There is an excess of 2.9 - 1.5 = 1.4 millimole of H+, in 15 ml of solution; the molar concentration is 1.4/15 = 0.093 M. pH=-log(0.093) = 1.03

NOTE: molar concentration (M) is moles/liter of solution same as millimoles / milliliter. Technically, the second result is approximate, since adding 0.06 of solid NaOH to 15 ml of solution will end up with more than 15 ml (but probably not much, so the approximation may be valid).

Answer 2

Well, first all the listed acids and bases here you should recognize as very strong acids (HCL, H2S04), and a very strong base (NaOH). Since very strong acids and bases will dissociate almost completely, there is no pKa or pKb for these compunds.
So, for question 1, find how many moles HCl there are (25.0ml) x ( 0.10mol/1000ml) = 2.5 x 10-3 = 0.0025 mol HCl. So, there are going to be 0.0025 H+ ions in solution. Also, find the moles of NaOH you are adding..(30ml) x (0.075mol/1000ml)= 0.0025 mol NaOH, and 0.0025 mol of OH- ions. So, you are adding the same number of moles of H+ and OH-. These should equalize each other, giving a neutral pH of 7. This is similar to a titration; when there is a one to one correspondance between HCl and NaOH (the same # of moles existing in solution) they reach an equivalance point, which for equally strong acids and bases in pH 7.


okay #2. There are 0.001 moles of HCl being added (10.0ml) x (0.1mol/1000ml) = 0.001 mol HCl. There is 9.5 x 10-4 moles of H2SO4 added : (5.0ml) x (0.19mol/1000ml) = 9.5 x 10-4 moles. Also, there are 0.0015 mol of NaOH added (moles = grams/amu); the amu for NaOH is 40, so moles NaOH = 0.06g/40 = 0.0015 mol NaOH.
Again, the NaOH and HCl are very close 0.001 to 0.0015, if you use rounding wisely, you can count then equal, so those two cancel out, and now you just need the pH of 9.5 x 10-4 mol H2SO4. pH = -log[H+] = -log[9.5 x 10-4] = 3.02.
A pH pf 3.02 is assuming that H2SO4, a diprotic acid will only give up one hydrogen atom which is usually the case since H2SO4 is such a stronger acid than HSO4-. However, just incase both protons are released, the pH = -log[0.0019] = pH 2.72.

Well, this is my best shot at it....I hope it helps!!!

Answer 3

1) Moles of HCl = 2.5 mmol

Moles of NaOH = 2.25 mmol

Thus, 0.25 mmol of HCl are left over

pH = -log(0.25 x 10^-3/55) = 5.342

2) HCl = 1 mmol
H2SO4 = 9.5 mmol which gives 19 mmol of H+
NaOH = 1.5 mmol

18.5 mmol of H+ are left

Answer 4

1) Moles of HCl = 2.5 mmol

Moles of NaOH = 2.25 mmol

Thus, 0.25 mmol of HCl are left over

pH = -log(0.25 x 10^-3/55) = 5.342

2) HCl = 1 mmol
H2SO4 = 9.5 mmol which gives 19 mmol of H+
NaOH = 1.5 mmol

18.5 mmol of H+ are left.

Answer 5

one problem w/ solution---
25mL of .10M HCl diluted w/ 100mL H2O what is pH? Then what will pH be of solution after 10 mL of .10 M NaOH is added???

Solution----
Part A-(0.025L)(.10 mol/L)===2.5x10^-3 mol so....
{H+}=(2.5x10^-3mol)/(.10 L)===2.5x10^-2 mol/L
pH is calculated by following---
pH=log(1/H+) OR log(1/2.5x10^-2)===pH=1.68

Part B- take original moles of NaOH (0.01 L)(0.10 mol/L)===0.001 mol so take this number away form moles of original moles of HCL (.0025 mol) so....0025-.001=======0.0015
{H+}=.0015 mol/(0.11 L)= 0.0136
pH can now be calculated
log(1/.0136)= 1.87===pH