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3 years ago

How many mL of 2.5M HCl would be needed to completely neutralize a standard solution of 0.53M NaOH in a titration

Chemistry

1 answer:

Elena L [17]3 years ago

4 0

Answer:

Amount of HCL = 0.00318 L of 3.18 ml

Explanation:

Given:

HCL = 2.5 M

NaOH = 0.53 M

Amount of NaOH = 15 ml = 0.015 L

Find:

Amount of HCL

Computation:

HCL react with NaOH

HCl + NaOH ⇒ NaCl + H₂O

So,

Number of moles = Molarity × volume

Number of moles of NaOH = 0.53 × 0.015

Number of moles of NaOH = 0.00795 moles

So,

Number of moles of HCl needed = 0.00795 mol es

So,

Volume = No. of moles / Molarity

Amount of HCL = 0.00795 / 2.5

Amount of HCL = 0.00318 L of 3.18 ml

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kvv77 [185]

Answer:

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3 years ago

A sample of seawater consists of water, sodium ions, chloride ions, and several other dissolved salts. These ions and salts are

stellarik [79]

Answer:

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