Select all that are true for a titration of 0.20 m khp with 0.20 m hcl as the titrant as compared to your titration with 0.20 m naoh as the titrant in the lab. (select all that apply.)
1. the equivalence volume would be the same.
2. the ph at the equivalence point would be the same.
3. phenolphthalein would be an appropriate indicator.
4. the initial ph would be the same.
5. the ph at the midpoint would be the same

1. The equivalence volume would be the same.

4. The initial pH would be the same.

Explanation:

The reactions of KHP with HCl and NaOH are:

KHP + HCl → H₂P + KCl

KHP + NaOH → P²⁻ + H₂O + K⁺ + Na⁺

As the molar concentration of HCl and NaOH are the same, 1. The equivalence volume would be the same.

The pH will not be the same because with HCl you are adding acid decreasing pH and for the second reaction NaOH is a base that increase pH. For the same reason, in the midpoint pH wouldn't be the same.

Phenolphtalein wouldn't be an appropriate indicator because the useful pH range is between 6,0 and 7,6 and the equivalence points are in 7,2 and 10,6.

4. The initial pH would be the same.  That's true because in the two titrations you will have just 0,20M KHP in solution.

I hope it helps!

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explanation:

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm