you start with a solution of 1.4 M HCl. you dilute 36 mL of this concentrated solution to a new volume of 1.5 L. what is the pH of this new, diluted solution of HCl?
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Let us see, should be straight forward.
You first find out the no. of moles of HCl in 36 mL or 0.036 L, After dilution they will be present in 1.5 L volume, then you find out the new molarity of HCl
No. of moles of HCl = 1.4 moles/L * 0.036 L = 0.0504 moles
New molarity = 0.0504 moles / 1.5 L = 0.0336 M
HCl has one H+, so [H+] = 0.0336 M
pH = - log [H+] = - log 0.0336 = 1.51
There you have it
You first find out the no. of moles of HCl in 36 mL or 0.036 L, After dilution they will be present in 1.5 L volume, then you find out the new molarity of HCl
No. of moles of HCl = 1.4 moles/L * 0.036 L = 0.0504 moles
New molarity = 0.0504 moles / 1.5 L = 0.0336 M
HCl has one H+, so [H+] = 0.0336 M
pH = - log [H+] = - log 0.0336 = 1.51
There you have it