1. Calculation the pH the a 0.01 M HCl solution. HCl is a solid electrolytes and thus its complete dissociation will develop a solution O.O1 M (10-2 M) in H+. Because the concentration that H+ arising from water dissociation (10-7 M) is especially lesser than 10-2 M, it have the right to be neglected and also thus pH = -Log (10-2) = 22. Calculation the pH the a 0.1 M NaOH solution.
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NaOH is a solid electrolytes and thus its finish dissociation will develop a equipment O.1 M (10-2 M) in OH-. By neglecting
Alternatively we can calculate pOH = - log in
remembering the definition of logarithms
By using the expression because that ion product the water:
Since NaOH is a strong base we need to add 5.01-14 mole of NaOH because that liter of water, 10.02-14 for two liters that water. By multiplying the moles v the molecular weight of NaOH we have(10.02-14) (40) = 0.04 grams
4. An HCl solution having actually pH = 3 should be diluted in together a method to acquire pH = 4. How plenty of water you should add?
remembering that M1V1 = M2V2(10-3)(3) = (10-4)(x)
x = (10-3)(3)/(10-4) = 30 litersThus you require to add 27 liters of water.
Alternatively because a 10-4 M equipment is tenfold much more diluted than a 10-3 M, the last volume must be(3)(10) = 305. 300 ml that a 0.02 M NaOH systems are combined with 200 ml the 10-2 M Ba(OH)2 solution. Calculation the pH of the resulting solution. Both the bases are totally dissociated in solution.The variety of moles of OH- coming from NaOH are:0.02 moles: 1000 ml = x mole : 300 ml x = (6) (10-3)The variety of moles that OH- coming from Ba(OH)2,considering that each mole the the basic produces 2 moles the OH-, are:
(2)(10-2) : 1000 = x : 200
x = (4)(10-3)
Thus the total number of moles that OH- is (6) (10-3) + (4)(10-3) = 0.01 and also since the final volume the the systems is 500 ml the solution results to be 0.02 M in OH-:
pOH = - log 0,02 = 1,69
pH =14-1,69 = 12,3
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Vinegar is a wine derivative comprise acetic mountain (pH = 3) and also flavoring agents. Vinegar can be simulated by prepare a solution of acetic acid having actually pH 3 and including appropiate flavoring agents. Calculate the ml that acetic mountain ( density = 1.049 g/ml, MW = 60, Ka = 1.74 * 10-5, pKa = 4.76) needed to prepare 1 liter the vinegar.
Assuming the the concentration of the mountain at equilibrium is equalto the early concentration that the mountain (Co) it has actually been demonstratedthat:pH = 1/2 (pKa - log in Co) = 1/2 pKa - 1/2 log in Coand solving this equation for Log Co we have:Log Co = (pH - 1/2pKa)*-1/2 = (3 -4.76/2)*-2
Log Co = (3 -2.38)*-2 = 0.62*-2 = -1.24Taking the antilogarithm the results:Co = 0.0575 M which synchronizes to 0.0575 * 60 (MW) = 3.42 gramsof acetic acid.From the value of density it outcomes that 3.42 grams correspondsto 3.45/1.049 = 3.28 ml the acetic acid .A better result can be derived considering the the concentration of acetic mountain calculated is the concentration at equilibrium not the early one. Because pH = 3 corresponds to 10-3 M H30+, the initial concentration of acetic mountain is : 0.057 + 10-3 = 0.058 M thus: ml = (0.058*60)/1.049 = 3.31