You have in front of you 100 mL of 7.00×10 2 M HCl, 100 mL of 5.00×10 2 M NaOH,

Question

You have in front of you

100 mL of 7.00×102M HCl,

100 mL of 5.00×102M NaOH, and

plenty of distilled water.

Part A

Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?

Express your answer to three significant figures and include the appropriate units.

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.60.

You have in front of you

100 mL of 7.00×102M HCl,

100 mL of 5.00×102M NaOH, and

plenty of distilled water.

You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 85.0 mL of HCl and 88.0 mL of NaOH left in their original containers.

Part A

Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

HCl used = 100-85 = 15 ml = 0.015 L

NaoH used = 100-88 = 12 ml = 0.012L

HCl moles present = Mx V = 7 x 10^-2 x 0.015 =0.00105

NaOH mole present = M x V = 5 x 10^-2 x 0.012 = 0.0006

HCl excess after getting neutralised by NaOH = 0.00105-0.0006 = 0.00045

we need pH = 2.6 , [H+] = 10^ -pH = 10^ -2.6 = 0.002512

H+ moles = M x V = 0.002512x 1L = 0.002512

we had 0.00045 moles Hl ( i.e moles of H+) , hence additional H+ moles needed to be added

exces H+ moles needed = 0.002512-0.00045 = 0.00206

Molarity of HCl = 7 x 10^ -2 M = 0.07 M

volume needed of HCl = moles / M = 0.00206 /0.07 = 0.02943 L = 29.43 ml

Thus we need 29.43 ml of moe HCl to original solution to get pH = 2.6

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