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expeople1 [14]
3 years ago
14

0.80 g of hydrogen chloride (hcl is dissolved in water to make 5.5 l of solution. what is the ph of the resulting hydrochloric a

cid solution?
Chemistry
1 answer:
scoundrel [369]3 years ago
8 0
HCl is a strong acid and so dissociates fully in water:
HCl(aq) --> H+(aq) + Cl-(aq)
Since there is a 1:1 mole ratio between HCl and H+ in this reaction, the number of moles of H+ = number of moles of HCl. The number of moles of HCl can be calculated as moles = mass/RMM = 0.80g / 36.5g/mol = 0.022mol.
Since 5.5 l  of solution is made, the concentration of H+ is:
[H+] = 0.022mol / 5.5l = 0.004M. This can then be used to calculate pH through the equation pH = -log[H+] = -log(0.004) = 2.40
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The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

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To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL

Putting values in above equation, we get:

2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL

Hence, the volume of NaOH needed is 36.2 mL

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