Adding witnesses to prove Gödel's completeness theorem I am currently working with "The Foundations of Mathematics" by Kunen to understand the proof for Gödel's completeness theorem due to Henkin.
When adding the witnessing terms, there is one thing I can't seem to understand.
First we show that given a set of sentences $\Sigma$ in the language $\mathscr{L}$, we may add $\delta$-many new constant symbols to $\mathscr{L}$. We call this new languag $\mathscr{L}_{\delta} := \mathscr{L}\cup \{c_{\alpha} | \alpha <\delta\}$. $\delta$ is an arbitrary ordinal. Now with the aid of these constants, we can add witnesses to $\Sigma$ ($\Sigma_{\delta}:=\Sigma \cup \{\exists x \varphi_{\alpha}(x) \rightarrow \varphi_{\alpha}(c_{\alpha}) | \alpha < \delta\}$) for the existential sentences $\exists x \varphi_{\alpha}(x)$ of $\mathscr{L}_{\alpha} := \mathscr{L}\cup \{c_{\xi} | \xi <\alpha\}$, s.t.
$$ Con_{\vdash,\mathscr{L}}(\Sigma) \Rightarrow Con_{\vdash,\mathscr{L}_{\delta}}(\Sigma_{\delta}) $$
This is Lemma II.12.19 in the book. Now $\Sigma_{\delta} $ has witnesses for the $\exists x \varphi_{\alpha}(x), \alpha < \delta$ but it may not have witnesses for all existential sentences of the new language $\mathscr{L}_{\delta}$, so we need to remedy this.
This is done with Lemma II.12.20 which states that for a consistent set of sentences $\Sigma$ in $\mathscr{L}$, you can actually find $\Sigma'$ and $\mathscr{L}'$, s.t. $\Sigma'$ has witnesses in all of $\mathscr{L}'$ and $Con_{\vdash,\mathscr{L}'}(\Sigma')$. Here we add $\kappa \cdot \omega$ new constants to $\mathscr{L}$, where $\omega$ is the first limit ordinal and $\kappa = max\{|\mathscr{L}|,\aleph_0\}$.
The part that I don't understand is why do we know that there are exactly $\kappa$ existential sentences of $\mathscr{L}_{\kappa \cdot n} := \mathscr{L}\cup \{c_{\alpha} | \alpha <\kappa \cdot n\}$?
Also I don't quite understand what that statement actually means. Does it say there are exactly $\kappa$ existential sentences that cannot be stated in $\mathscr{L}_{\kappa \cdot (n-1)}$ and don't need symbols of $\mathscr{L}_{\kappa \cdot (n+1)} \setminus \mathscr{L}_{\kappa \cdot n}$?
How can we determine how many existential sentences can be stated in a language? Especially if we do not know the language.
For example if we consider the language of ordered rings $\mathscr{L} = \{1,0,<,+,-,\cdot\}$, we may state a sentence of the form $\exists x (x < 1+\dots +1)$ with $n$ ones being added up for every $n < \omega$. 
However, if we look at $\mathscr{L}=\{<\}$, we could not do that as we neither have the addition $+$ nor the constant $1$ at hand.
Any help would be very much appreciated. Thank you :)
 A: 
The part that I don't understand is why do we know that there are exactly $\kappa$ existential sentences of $\mathscr{L}_{\kappa \cdot n} := \mathscr{L}\cup \{c_{\alpha} | \alpha <\kappa \cdot n\}$?

We know there are at most $\kappa$ existential sentences, because each existential sentence is a finite sequence of symbols from an alphabet consisting of the basic logical symbols, a countable infinity of variable letters, the symbols of $\mathscr L$, and the $\kappa\times_c n=\kappa$ new constant letters. This alphabet has size $\kappa$, so the number of existential sentences is no larger than
$$ \sum_{n\in\omega} \kappa^n = \sum_{n\in\omega} \kappa = \omega\times_c\kappa = \max(\omega,\kappa) = \kappa $$
(This is basic cardinal arithmetic).
On the other hand, there are also at least $\kappa$ existential sentences, even before we begin using the new constant letters. For example, there are $\aleph_0$ formulas of the form $\exists x(x=x\land x=x\land \cdots \land x=x)$. This is already enough, unless $\mathscr L$ is uncountable such that $\kappa>\aleph_0$. But in that case we can also take $\exists x\,p(x,\ldots, x)$ for each predicate letter $p\in\mathscr L$, $\exists x(x=f(x,\ldots,x))$ for each function letter $f\in\mathscr L$, and $\exists x(x=c)$ for each constant letter $c\in\mathscr L$.
Cantor-Schröder-Bernstein now gives that there are exactly $\kappa$ existential sentences.
In fact, however, the proof would still work well enough if there were fewer than $\kappa$ existential sentences; just choose some of the $\varphi_\alpha$s to be equal! We need a witness for each existential sentence, but it doesn't harm anything if, for lack of enough properties to witness, some of the existential sentences get more than one witness constant.

Also I don't quite understand what that statement actually means. Does it say there are exactly $\kappa$ existential sentences that cannot be stated in $\mathscr{L}_{\kappa \cdot (n-1)}$ and don't need symbols of $\mathscr{L}_{\kappa \cdot (n+1)} \setminus \mathscr{L}_{\kappa \cdot n}$?

That's not what it actually says (and not what is needed; see above remarks about reusing $\varphi_\alpha$s), but it may be most intuitive to imagine that.
In that case you would need to show that there are at least $\kappa$ sentences that can be made with $\mathscr{L}_{\kappa\cdot n}$ but not with $\mathscr L_{\kappa\cdot(n-1)}$. But the sentences $\exists x(x=c_\alpha)$ for every $\alpha$ between $\kappa\cdot(n-1)$ and $\kappa\cdot n$ will do fine for that.

Especially if we do not know the language.

As long as the language allows you to form $\exists x(x=c_\alpha)$ for each different $c_\alpha$, there will be enough sentences. The size of $\mathcal L$ doesn't matter for this -- except if there are more than countably many symbols in $\mathscr L$, in which case they're going to push $\kappa$ itself upwards.
If $=$ is not a primitive logical symbol for you, it needs to be tacitly assumed that $\mathscr L$ does contain at least one predicate letter of nonzero arity, such that every term can be put into a wff somehow. But if it doesn't, then the completeness theorem holds vacuously, because everything collapses into propositional logic at best.
