- log(0.0235 M HCl)
= 1.6 pH
=============If I remember correctly two places are a pH designation standard.
D) 5
pH = 4. The reason is that pH = -log [H+] or -log [H3O+] = -log 10^-4 = 4.
100 Liters? I will assume as much. Molarity = moles of solute/Liters of solution Molarity = 0.10 mole HCl/100.0 Liters = 0.001 M HCl -------------------------now, to find pH - log(0.001 M HCl) = 3 pH -----------------so, your acid is of 3 pH, which is to be expected at the volume od solution
Given: 27 mL of NaOH, 0.45M; 20 mL HCI Need: M of HCI 27 ml NaOH*(1 L NaOH/1000mL NaOH)*(0.45M NaOH/1L NaOH)*(1mole HCI/1 mole NaOH)=0.012 0.012/0.02=0.607 M HCI (or rounded 0.61 M HCI)
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
D) 5
pH = 4. The reason is that pH = -log [H+] or -log [H3O+] = -log 10^-4 = 4.
A solution of HCl is highly dissociated into ions, A 0.000001 M solution (1 x 10-6) has a pH of 6 ... close to neutral. A 0.001 M solution (1 x 10-3) has a pH of 3 ... more concenterated, but still not a really concentrated solution. A 0.1 M solution (1 x 10-1) has a pH of 1 ... even more concentrated. showing it is more acidic.
100 Liters? I will assume as much. Molarity = moles of solute/Liters of solution Molarity = 0.10 mole HCl/100.0 Liters = 0.001 M HCl -------------------------now, to find pH - log(0.001 M HCl) = 3 pH -----------------so, your acid is of 3 pH, which is to be expected at the volume od solution
Given: 27 mL of NaOH, 0.45M; 20 mL HCI Need: M of HCI 27 ml NaOH*(1 L NaOH/1000mL NaOH)*(0.45M NaOH/1L NaOH)*(1mole HCI/1 mole NaOH)=0.012 0.012/0.02=0.607 M HCI (or rounded 0.61 M HCI)
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
its PH is 3
A molarity of 25? I think you have made some mistake. Please ask this question again.
- log(0.12 M) = 0.92 pH ---------------
-log(0.5 M HF) = 0.3 pH
- log(0.000626 M H2SO4) = 3.2 pH -----------
A 0.01 M solution of NaOH has a pH =13