The molarity of the lake water is 0.00001 M and the pH of lake water is 5.
The lake water is acidic.
Explanation:
Data given:
molarity of base solution Mbase = 0.1 M
volume of the base solution Vbase = 0.1 ml or 0.0001 litre
volume of lake water Vlake = 1000ml or 1 litre
molarity of the lake water, Mlake = ?
Using the formula for titration:
Mbase X Vbase = Mlake X
Mlake = 
Putting the values in the equation:
Mlake = 
Mlake = 0.00001 M
The pH of the lake water will be calculated by using the following formula:
pH = - 
[
]
pH = - [ 0.00001]
pH = 5
Answer:
pH of resulting solution = 7.98
Explanation:
The balanced equation
HA + NaOH - Na+ + A- + H2O
Number of moles of A = Number of moles of HA = Number of moles of NaOH
= 35.8/1000 * 0.020 = 0.000716 mol
Initial concentration of A = 0.000716/0.0608 = 0.01178 M
pKb = 14 – pKa = 14 -3.9 = 10.1
Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11
Kb = [HA][OH-]/[A-]
Kb = a^2/(0.01178 -a) = 7.943 * 10^-11
a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0
a = 9.673 * 10^-7
OH- = a = 9.673 * 10^-7 M
pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02
pH = 14-6.02 = 7.98
Answer:me being a tanned ginger
Explanation: i cant date u
Marine west coast, humid subtropical, and Mediterranean.
Answer:
70.1 mL
Explanation:
First let's look at the formula for magnesium nitrate, and get the molar mass, we should end up with Mg(NO3)2 for the formula and this should have a molar mass of 148.3 g/mol.
Lets get the number of moles of the magnesium by taking the number of grams over the molar mass, (21.94 g)/(148.3 g/mol). grams cancel and we're left with approximately 0.148 moles.
Now let's plug our numbers into the molarity formula, M = n/L, this should give us 2.11 mol/L = (0.148 mol)/L, now let's solve for L, divide both sides by 0.148 which will give us 14.26 L^-1 = 1/L now we take the inverse of both sides to get 0.07012 L = L.
Now we have the liters, but the question askes for milliliters, so let's multiply by 1000, and then after rounding to sig figs we will get 70.1 mL as our answer. (Note: I used the exact values instead of the approximations throughout this explanation, so if you calculate the answer by plugging in these values, it might be slightly off.)
