pH = 1
HCl is a strong acid so fully dissociates, so the concentration of H+ is equal to that of the HCl and pH=-log[H+]
Take the minus log value of 0.00001.pH equals to five.pH+pOH=14 pOH=9
0.1 molar means hydrogen ion concentration is 10^-1.
so pH=-log[H+] ion.
so pH =1.
again we know that ,
pH+poh=14
so poh =14-1=13
pH = -log [H3O+] = -log 0.1M = 1
pOH + pH = 14
Thus, pOH = 14 - pH = 14 -1 = 13
pOH= -log[OH]
1
Molarity is the concentration of a solution, defined as moles per unit volume. Where, Molarity = moles / volume In this case the molarity of the HCl solution is 0.03 M The pH of this is calculated by the equation below pH = - log [H+] Where [H+] is the concentration of hydrogen ions/ protons present in the solution. As HCl only contains one hydrogen ion per molecule then the concentation of [H+] is 0.03 M Then the equation can be calculated (the minus sign is very important!) pH = - log 0.03 pH = 1.52 To summarise, Molarity of 0.03 M HCl solution is 0.03 M pH of 0.03 M HCl solution is 1.52
- log(0.00450 M HCl)= 2.3 pH=======
pH= -log[H+] pH + pOH = 14 pOH = 14 - pH pOH= -log[OH], so the antilog of -pOH will give you the OH concentration.
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602
pH + pOH = 14. So pOH = 14 - 6 = 8 pOH = -log[OH-] [OH-] = 10-8 M
Molarity is the concentration of a solution, defined as moles per unit volume. Where, Molarity = moles / volume In this case the molarity of the HCl solution is 0.03 M The pH of this is calculated by the equation below pH = - log [H+] Where [H+] is the concentration of hydrogen ions/ protons present in the solution. As HCl only contains one hydrogen ion per molecule then the concentation of [H+] is 0.03 M Then the equation can be calculated (the minus sign is very important!) pH = - log 0.03 pH = 1.52 To summarise, Molarity of 0.03 M HCl solution is 0.03 M pH of 0.03 M HCl solution is 1.52
- log(0.00450 M HCl)= 2.3 pH=======
pOH= -log(0.0220M OH) pH=14-pOH pH=12.34242268
pH= -log[H+] pH + pOH = 14 pOH = 14 - pH pOH= -log[OH], so the antilog of -pOH will give you the OH concentration.
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602
pH + pOH = 14. So pOH = 14 - 1.12 = 12.88 pOH = -log[OH-] [OH-] = 1.31 x 10-13 M
pH + pOH = 14. So pOH = 14 - 6 = 8 pOH = -log[OH-] [OH-] = 10-8 M
- log(0.25 M HCl) = 0.6 pH ------------
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
[OH-] = 3.31 log[OH-] = pOH = .51982 14-pOH = pH = 13.48
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
pH=-lg[H+][H+]=10-pHWith pH=2.0:Corrected:[H+]= 10-pH = 10-2.0 = 0.010 M HCL